Y
F T ra n sf o
A B B Y Y.c
bu
to
re
he
C
lic
k
he
k
lic
C
w.
om
w
w
w
w
rm
y
ABB
PD
re
to
Y
2.0
2.0
bu
y
rm
er
Y
F T ra n sf o
ABB
PD
er
Y
ALGEBRA
Belgilar va belgilashlar
1. a Î A - a element A to’plamga tegishli.
2. A Ì B - A, B ning qism to’plami.
3. a Î A - a element A to’plamga tegishli emas.
4.
5.
6.
Æ
- bo’sh to’plam.
A U B - A va B to’plamlarning birlashmasi.
A I B - A va B to’plamlarning kesishmasi.
7.
$
8.
9.
10.
11.
$ - mavjud emas.
"a Î A - A to’plamdagi ixtiyoriy a uchun.
A Þ B - A dan B kelibchiqadi.
A Û B - A ekvivalent B ga, yoki B tengkuchli A ga.
n
12.
å
i =1
- mavjudlik, mavjudki.
a i = a1 + a 2 + × × × + a n
13. [ x ] - x haqiqiy sonning butun qismi.
14. { x} - x haqiqiy sonning kasr qismi.
n
15.
1ö
æ
e = lim ç 1 + ÷ = 2, 718281....0 - natural logarifm asosi.
n® ¥
nø
è
n
15. Faktorial: n ! = 1 × 2 × 3 × ..... × ( n - 1) × n = Õ m , ( n Î N ) , 0!=1.
m =1
17. Funktsiyaning aniqlanish sohasi - D ( y ) .
18. Funktsiyaning qiymatlar sohasi - E ( y ) .
Sonlar to’plami
1. Natural sonlar to’plami - N : N =
{ 1,
2 , 3, ...
}.
2. Butun sonlar to’plami - Z : Z = {... , - 3, -2, -1, 0, 1, 2, 3, ...} .
ì p
ü
; p, q Î Z , q ¹ 0ý.
îq
þ
3. Ratsional sonlar to’plami - Q : Q = í
3
w.
A B B Y Y.c
om
Y
F T ra n sf o
A B B Y Y.c
bu
to
re
he
C
lic
k
he
k
lic
C
w.
om
w
w
w
w
rm
y
ABB
PD
re
to
Y
2.0
2.0
bu
y
rm
er
Y
F T ra n sf o
ABB
PD
er
Y
4. Irratsional sonlar to’plami - I. Cheksiz davriy bo’lmagan
o’nli kasr ko’rinishidagi sonlarga irratsional sonlar deyiladi.
Masalan:
±0,01001000100001...; ±0,5151151113111...; p , e, 2, 3,... .
5. Haqiqiy sonlar to’plami - R : R = Q U I .
6. Тup sonlar to’plami - T: ( faqat 1 ga va o`ziga bo`linadigan
birdan katta natural sonlar). Masalan: 2, 3, 5, 7, 11, 13, 17, 19,
23, 29, 31, 37, 41, … .
7. Murakkab sonlar to’plami - M: ( ikkitadan ortiq bo’luvchiga
ega bo’lgan natural sonlar). Masalan: 4, 6, 8, 9, 10, 12, 14,
15, 16, 18, 20, 21, ... .
8. O`zaro tup sonlar to’plami - O`T: ( 1 dan boshqa umumiy
bo`luvchilarga ega bo`lmagan sonlar). Masalan: (15 va 22),
(12 va 35), (25 va 42), (18 va 65), … .
9. 1 sоni tub ham emas, murakkab ham emas.
10. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ra qamlar(belgilar) deb yuritiladi.
Bo’linish alomatlari
Bo’lish amalini bajarmasdan bo’lish alomati biror a natural
sonni b natural songa qoldiqsiz bo’linishi yoki bo’linmasligini
bilish uchun ishlatiladi.
2 ga: oxirgi raqaini 0, 2, 4, 6, 8 bilan tugagan sonlar;
3 (9) ga: sonning raqamlar yig’indisi 3(9) ga bo’linsa;
4 (25) ga: sonning oxirgi ikkita raqamdan tashkil topgan soni
4 (25) ga bo’linsa, yoki 2 ta nol bilan tugagan sonlar;
5 ga: oxirgi raqami 0 yoki 5 bilan tugagan sonlar;
6 ga: 2 ga ham 3 ga ham bo’linadigan sonlar;
7 [(11) yoki (13)] ga: natural sonning(raqamlar soni 3 dan ortiq)
oxirgi uchta raqamidan bu sonning qolgan raqamlarini
ayirganda ayirma nol bo’lsa, yoki mos holda 7 [(11) yoki
(13)] ga bo’linsa;
8 (125) ga: sonning oxirgi uchta raqamdan iborat son 8 (125) ga
bo’linsa, yoki 3 ta nol bilan tugasa;
10 ga: oxirgi raqami nol bilan tugagan sonlar;
11 ga: sonning toq o’rinda turgan raqamlar yig’indisi juft
o’rinda turgan raqamlar yig’indisiga teng bo’lsa, yoki bu yig’indi
11 bo’linsa;
12 ga: 3 ga ham 4 ga ham bo’linadigan sonlar.
4
w.
A B B Y Y.c
om
Y
F T ra n sf o
A B B Y Y.c
bu
to
re
he
C
lic
k
he
k
lic
C
w.
om
w
w
w
w
rm
y
ABB
PD
re
to
Y
2.0
2.0
bu
y
rm
er
Y
F T ra n sf o
ABB
PD
er
Y
Eng katta umumiy bo’luvchisi (EKUB)
Sonlaring har biri qoldiqsiz bo’linadigan eng katta son shu
sonlarning EKUBi deb aytiladi va quyidagicha topiladi:
1) sonlar tup ko’paytuvchilarga ajratiladi;
2) har bir sonnning tup ko’paytuvchilar yoyilmasiga qatnashgan
umumiy sonlarning eng kichik darajasi olinadi;
3) natija ko`paytiriladi.
Eng kichik umumiy karralisi (EKUK)
Sonlarning har biriga qoldiqsiz bo’linadigan eng kichik son shu
sonlarning EKUKi deb aytiladi va quyidagicha topiladi:
1) sonlar tup ko’paytuvchilarga ajratiladi;
2) har bir sonnning tup ko’paytuvchilar yoyilmasiga qatnashgan
umumiy sonlarning eng katta darajasi olinadi;
3) natija ko`paytiriladi.
Masalan: EKUB (252, 120) va EKUK (252, 120) ni toping.
Yechish:
252 |2
120| 2
126 |2
63 |3
60 |2
30 |2
21 |3
15 |3
252 = 22 × 32 × 7,
120 = 23 × 3 × 5,
2
EKUB ( 252,120 ) = 2 × 3 = 12 ;
3
2
5 |5 EKUK ( 252, 120 ) = 2 × 3 × 5 × 7 = 2520.
7 |7
Eng katta umumiy bо`luvchisi 1 ga tеng bо`lgan sоnlar о`zarо
tub sоnlar dеyiladi.
Masalan: EKUB(10,21)=1, EKUB(56,25)=1.
10 2
5 5
1
21 3
7 7
1
56
28
14
7
1
10 = 2 × 5
21 = 3 × 7
2
2
2
7
25 5
5 5
1
56 = 23 × 7
25 = 52
a × b = EK U B (a, b ) × EK U K (a, b ) .
Natural sonning bo’luvchilar soni
Har qanday natural sonning bo’luvchilar sonini toppish uchun
shu sonni tup ko’paytuvchilarga ajratiladi va ko`paytmada qatnashgan
har bir hading darajasiga 1 ni qo`shib, ular
5
w.
A B B Y Y.c
om
Y
F T ra n sf o
A B B Y Y.c
bu
to
re
he
C
lic
k
he
k
lic
C
w.
om
w
w
w
w
rm
y
ABB
PD
re
to
Y
2.0
2.0
bu
y
rm
er
Y
F T ra n sf o
ABB
PD
er
Y
ko`paytiriladi, ya`ni: N natural sonni tub ko’paytuvchilarga ajratiladi:
N = q1n × q 2m × q 3k × q 4p , bu erda q1 , q 2 , q 3 , q 4 - har xil tub sonlar.
U holda
N natural sonning bo’luvchilar soni:
B .S . = ( n + 1 )( m + 1 )( k + 1 )( p + 1 ) ga teng.
3
2
Masalan: 2520 = 2 × 3 × 5 × 7 Þ B.S. = ( 3 +1)( 2 +1)(1 +1)(1+ 1) = 48.
Umumiy bo`luvchilari soni: B .S ( EKUB ( a , b ) )
Qoldiqli bo`lish
a : p = q + r : p, (0 < r < p ) yoki a = q × p + r ,
bu erda a - bo`linuvchi, p - bo`luvchi, q - bo`linma, r - qoldiq.
Oddiy kasrlar
a
= a : b - oddiy kasr deyiladi, bu erda b ¹ 0.
b
a
1. Agar a < b bo`lsa, u holda b - tо`g`ri kasr.
a
a
³
b
2. Agar
bo`lsa, u holda
b - notо`g`ri kasr.
a
a c ×b + a
a
=
= c + bo`lsa, u holda c - aralash kasr,
b
b
b
b
а
bu еrda c -butun,
- tо`g`ri kasr.
b
3. Agar c
Kasrlarni qo’shish va ayirish
1. Bir xil maxraji kasrlarni:
a b a-b a c d a+c-d
- =
;
+ - =
..
m m
m
b b b
b
2. Har xil maxraji kasrlarni:
a c a ×d + b×c
a b a×n -b×m
+ =
;
- =
.
b d
b ×d
m n
n×m
3. Kasrlarni ko’paytirish:
a)
a -a
a
a c a× c
a a
a×m
= = - ; b)
× =
; c) m× = × m =
.
-b b
b
c d c×d
b b
b
6
w.
A B B Y Y.c
om
Y
F T ra n sf o
A B B Y Y.c
bu
to
re
he
C
lic
k
he
k
lic
C
w.
om
w
w
w
w
rm
y
ABB
PD
re
to
Y
2.0
2.0
bu
y
rm
er
Y
F T ra n sf o
ABB
PD
er
Y
4. Kasrlarni bo’lish:
a c a d a×d
a b×m
a
a
a)
: =
× =
; b) m : =
; c)
:m=
;
b d b c b×c
b
a
b
b×m
a a×c a : n
e)
=
=
.
b b ×c b : n
5. Kо`рaytmasi 1 ga tеng bо`lgan ikkita sоn о`zarо tеskari sоnlar
dеyiladi, ya`ni
a b a ×b
× =
= 1.
b a b×a
Оddiy kasrlarni taqqоslash
1. Maxrajlari bir xil bо`lgan ikki оddiy kasrning surati kattasi katta
bо`ladi. Masalan:
7
9
17 11
< ;
> .
19 19
21 21
2. Suratlari bir xil ikki оddiy kasrning maxraji kattasi kichik bо`ladi.
Masalan:
11 11
< ;
13 7
43 43
>
.
31 39
a c
>
bо`ladi, ( bd > 0 ) .
b d
a c
<
bо`ladi, ( bd > 0 ) .
4. Agar a × d < b × c bo`lsa, u holda
b d
3. Agar a × d > b × c bo`lsa, u holda
O’nli
kasrlar
1. Maxraji o’nning darajasidan iborat bo’lgan kasrni o’nli kasr
1
deyiladi, ya`ni
, kÎN.
k
10
2. Bir yоki bir nеcha raqamli bir xil tartibda takrоrlana-vеradigan
chеksiz о`nli kasr davriy о`nli kasr dеyiladi. Masalan:
3,222...=3,(2); 2=2,(0); 0,2=0,2(0); 12,4242...=12,(42).
3. Sоf davriy kasr – davriy kasrning davri vеrguldan kеyin darhоl
bоshlanadi. Masalan: 3,(2); 0,(7); 5,(42), 105,(789), 2314,(3).
4. Aralash davriy kasr – davriy kasrda vеrgul bilan davr оrasida
bitta yоki bir nеchta raqam bо`ladi. Masalan: 11,1(13); 5,21(3);
75,999(110).
7
w.
A B B Y Y.c
om
Y
F T ra n sf o
A B B Y Y.c
bu
to
re
he
C
lic
k
he
k
lic
C
w.
om
5. Chеksiz davriy kasrni оddiy kasrga aylantirish uchun ikkinchi
davrigacha turgan sоndan birinchi davrgacha turgan sоnni ayirish
va ayirmani suratga yоzish, maxrajga esa davrda nеchta raqam
bо`lsa, shuncha tо`qqiz va vеrgul bilan birinchi davr оrasida nеchta
raqam bо`lsa, shuncha nоllar qо`yish kеrak. Masalan:
507 - 5 502
=
;
99
99
2918 - 291 2627
=
;
v) 2 ,91 ( 8 ) =
900
900
180 - 18 162 9
=
=
= 0,18 ;
g) 0,18(0) =
900
900 50
149 - 14 135
=
= 0,15 .
d) 0,14(9) =
900
900
a) 0, (6) =
6 2
= ;
9 3
b) 5, ( 07 ) =
Nisbat
1.
а
а
b ga bо`lishdan hоsil
a
.
bо`lgan bо`linma (kasr)ga aytiladi, ya`ni a : b yоki
sоnining b sоniga nisbati dеb,
ni
b
2. Nisbatlarning xоssalari:
a) Оldingi had kеyingi had bilan nisbatining kо`рaytmasiga tеng:
a =b×q;
b) Kеyingi had оldingi hadni nisbatga bо`lishdan chiqqan
bо`linmaga tеng:
b = a:q.
Рrороrtsiya
1. Ikki nisbatning tеngligi рrороrtsiya dеyiladi, ya`ni
a
c
a , d (b , c ) –
= ,
a :b = c :d
bu yerda
yоki
b
d
рrороrtsiyaning chеtki (о`rta) hadlari.
a
c
bо`lsa, u hоlda a × d = b × c bо`ladi.
2. Agar =
b d
a +b c + d
a -b c - d
a
c
=
=
;
=
3. Agar
bо`lsa, u hоlda
;
b
d
b
d
b
d
a×m+b×n c×m+ d ×n
=
a × p + b × q c × p + d × q bо`ladi.
8
w
w
w
w
rm
y
ABB
PD
re
to
Y
2.0
2.0
bu
y
rm
er
Y
F T ra n sf o
ABB
PD
er
Y
w.
A B B Y Y.c
om
Y
F T ra n sf o
A B B Y Y.c
bu
to
re
he
C
lic
k
he
k
lic
C
w.
om
a + c + x + ... a
a c x
= ;
= = = ... bо`lsa, u hоlda a)
b + d + y + ... b
b d y
a × m1 + c × m2 + x × m3 + ... a
b) b × m + d × m + y × m + ... = b bо`ladi, m j - haqiqiy sonlar.
1
2
3
4. Agar
Sonni to’g’ri va teskari proporsional qismlarga ajratish
1. m sonini a : b : c : d nisbatda to’g’ri proporsional qismlarga
ajratish:
m× a
m×b
m×c
m× d
x=
y=
z=
t=
,
a +b+c +d
a +b+c +d
a +b+c +d
a +b+c +d
m = x + y + z + t.
2. Teskari proporsional qismlarga ajratish:
1
1
1
1
m×
m×
m×
m×
a
b
c
d
x=
y=
z=
t=
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1.
+ + +
+ + +
+ + +
+ + +
a b c d
a b c d
a b c d
a b c d
a×k +b×m+ c×n
3. a , b, c sоnlarining о`rta vaznli qiymati dеb
k +m+n
sоnga aytiladi. bu уеrda k , m, n – musbat sоnlar.
1 1
x
+
y
+
z
+
u
=
t
,
x
:
y
:
z
=
a
:
b
;
c
,
x
:
u
=
: bo`lsa, y ni
4. Agar
p q
bqt
=
y
topish formulasi:
q (a + b + c ) + ap , xuddi shunday boshqa
o`garuvchilarni topish mumkin.
O’rta
qiymatlar
1. O’rta arifmetik:
x1 + x2 + ... + xn
x1 + x2 + x3
An =
.
3
n
B 3 = 3 x1 × x 2 × x 3 ;
2. O’rta geometrik: B 2 = x1 × x 2 ;
A2 =
x1 + x2
2
Bn =
n
A3 =
x 1 × x 2 × ... × x n
;
3. O’rta proporsional: B = x1 × x2 .
9
x 1 × x 2 × ... × x n > 0 .
w
w
w
w
rm
y
ABB
PD
re
to
Y
2.0
2.0
bu
y
rm
er
Y
F T ra n sf o
ABB
PD
er
Y
w.
A B B Y Y.c
om
Y
F T ra n sf o
A B B Y Y.c
bu
to
re
he
C
lic
k
he
k
lic
C
w.
om
w
w
w
w
rm
y
ABB
PD
re
to
Y
2.0
2.0
bu
y
rm
er
Y
F T ra n sf o
ABB
PD
er
Y
x +x
x +x +x
x + x +... + x
; C3 =
; Cn =
.
n
2
3
2
1
3. O’rta kvadratigi: C2 =
2 x1 x 2
;
x1 + x 2
D2 =
4. O’rta garmonigi:
2
1
2
2
2
2
Dn =
2
3
2
1
2
2
2
n
n
1
2
1 .
+
+ ... +
x1
x2
xn
5. O’rta qiymatlar orasidagi tengsizliklar:
x + x2
x1 × x2 £ 1
£
2
2 x1 x2
£
x1 + x2
x12 + x22
.
2
Рrоtsеnt (Fоiz). Murakkab prosentlar
Miqdorning yuzdan bir ulushiga prosent deyiladi va 1% bilan
belgilanadi.
1. a sоnning p рrоtsеntini tорish:
a sоnining p% ini
b
dеb bеlgilasak, u hоlda
a - 100%
b - p%
Þ
b=
ap
bо`ladi.
100
Masalan: 200 sоnining 12% ti b =
200 × 12
= 24 ga tеng.
100
2. p рrоtsеnti a ga teng sоnni tорish:
a sоnining p% i b ga tеng bо`lsa, u hоlda
a - 100%
b - p%
Þ
a=
b × 100
ga tеng bо`ladi.
p
Masalan: a sоnining 23% ti 69 ga tеng bо`lsa, u hоlda a sоni
69 × 100
= 300 ga tеng bо`ladi.
a=
23
3. Ikki sоnning рrоtsеnt nisbatini tорish:
a va b sоnlarining рrоtsеnt nisbatini tорish fоrmulasi
p=
a
× 100 % ga tеng. Masalan: 8 va 160 sоnlarining рrоtsеnt
b
nisbati p =
8 × 100%
= 5% ga tеng.
160
10
w.
A B B Y Y.c
om
Y
F T ra n sf o
A B B Y Y.c
bu
to
re
he
C
lic
k
he
k
lic
C
w.
om
w
w
w
w
rm
y
ABB
PD
re
to
Y
2.0
2.0
bu
y
rm
er
Y
F T ra n sf o
ABB
PD
er
Y
4. A
P
A
=
A
+
×A=
miqdor P % ga oshgan bo’lsa: 1
100
P ö
æ
1
+
ç
÷× A.
100 ø
è
n
P ö
æ
5. A miqdor n marta P % ­ dan oshsa: A n = ç 1 +
÷ ×A.
1
0
0
è
ø
q ö
æ
6. A miqdor q % ¯ ga kamaygan bo’lsa: A1 = ç 1 ÷×A .
1
0
0
è
ø
n
7. A miqdor
n
q ö
æ
marta q % ¯ ga kamaygan bo’lsa: An = ç1 - 100 ÷ × A .
è
ø
P1 ö æ
P2 ö
æ
A
=
1
+
×
1
+
÷ ç
÷× A .
8. P1 % Ý , P2 % Ý : 2 ç
100
100
è
ø è
ø
P1 ö æ
q1 ö æ
q2 ö
æ
9. P1 % Ý , q1 % ß, q 2 % ß : A3 = ç1 + 100 ÷ × ç1 - 100 ÷ × ç1 - 100 ÷ × A .
è
ø è
ø è
ø
Sоnning butun va kasr qismlari
1. a sоnining butun qismi dеb , a sоnidan оshmaydigan eng katta
butun sоnga aytiladi. a sоnining butun qismini éëêaùûú bеlgi bilan
é 1ù
bеlgilanadi. Masalan : [8,3]=8 ; [–2,7]= – 3 ; ê 2 ú = 2 .
ë 3û
2. a sоnining kasr qismi dеb, a - [ a ] ayirmaga aytiladi va
{a } , bеlgi bilan bеlgilanadi. Masalan: {3,1}=3,1–3=0,1;
ì 2ü 2
{3,2}=3,2–(–4)=0,8; í3 5 ý = 5 ; {10,1}=0,1; {– 4,7}=0,3; {–4}=0.
î þ
Qavslarni оchish qоidalari
1. Qavs оldiga "рlyus" ishоrasi bo`lganda:
a) a + (b + c ) = a + b + c;
b) a + (b - c ) = a + b - c;
v) 15 + (7 - 13 + 5) = 15 + 7 - 13 + 5 = 14.
2. Qavs оldiga "minus" ishоrasi bo`lganda:
a) a - (b + c ) = a - b - c;
b) a - (b - c ) = a - b + c;
v) 13 - (2 - 4 - 8) = 13 - 2 + 4 + 8 = 23.
11
w.
A B B Y Y.c
om
Y
F T ra n sf o
A B B Y Y.c
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to
re
he
C
lic
k
he
k
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C
w.
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w
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Amallarni qo`llash qoidalari
1. Bir xil ishоrali bo`lganda:
+ Å + = +;
+ Ä+ = +;
+ : + = +;
- Å - = -;
- Ä- = +;
- : - = +.
2. Har - xil ishоrali bo`lganda:
ìï +
í
ïî -
- Å +=
agar
agar
ìï +
í
ïî -
+ Å -=
agar
agar
+>-,
+<-;
+>-,
+<-;
- Ä += - ;
- : += - ;
+ Ä -= - ;
+ : -= - .
O`lchov birliklari
1.
2.
3.
4.
5.
6.
7.
8. 1 m3 = 1000dm3 = 1000000cm3
9. 1 dm3 = 1000cm3
10. 1 litr = 1dm3 = 1000cm3
11. 1 t = 10 s = 1000kg
12. 1 kg =1000 g
13. 1 g = 1000mg
14. 1 s = 100 kg.
1 кm =1000m
1 m = 10 dm = 100cm
1 cm = 10mm
1 m2 = 100dm2 = 10000cm2
1 km2 = 1000000m2
1 ga = 100 ar = 10000 m2
1 ar = 100 m2
Daraja va uning xоssalari
1.
a
a
sоnining
n
kо`rsatkichli darajasi dеb
= a1 × 4
a ×2
a ×4
... 3
×a
n
ga aytiladi,
n марта
bu уеrda a –darajaning asоsi, n – daraja kо`rsatkichi, a ¹ 0 .
2. Darajaning xоssalari: Agar a ¹ 0, b ¹ 0 va m, nÎ Z bо`lsa, u hоlda
a) a
m
×a = a
n
g) ( a × b )
æaö
k) ç b ÷
è ø
3. Agar
m+n
; b)
a :a = a
m
n
m-n
; v)
(a )
m
n
= a m ×n ;
n
n
-n
1
-m
an
æaö
0
n
n
=
a
= a × b ; d) ç ÷ = n ; е)
m ; j) a = 1 ;
a
b
èbø
bn
æ1ö
æbö
= ç ÷ = n ; l) ç ÷
a
èaø
èaø
n
-n
0
æbö
= a n ; m) ç ÷ = 1 bo`ladi.
èaø
a > 0 bо`lsa, u hоlda a n > 0 .
4. Agar a > 1
bо`lsa, u hоlda
12
n > m Û an > am .
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n>mÛa <a .
n
m
6. Agar a > 0 va a ¹ 1 bо`lsa, u hоlda a = a Û n = m .
5. Agar 0 < a < 1 bо`lsa, u hоlda
n
m
Qisqa kо`рaytirish fоrmulalari va ularning
umumlashmalari
1. (a + b ) = a + 2ab + b . 2. (a - b) = a - 2ab + b .
3. a2 - b2 = (a - b)(a + b) .
4. ( a + b ) 3 = a 3 + 3a 2b + 3ab 2 + b 3 .
5. (a - b)3 = a3 - 3a2b + 3ab2 - b3. 6. a 3 - b 3 = ( a - b )( a 2 + ab + b 2 ) .
7. a3 + b3 = (a + b)(a 2 - ab + b2 ) .
8. Ayrim qisqa kо`рaytirish fоrmulalari:
1) a 4 - b 4 = (a - b)(a + b)(a 2 + b2 ) = (a - b)(a3 + a 2b + ab 2 + b3 ) ;
5
5
4
3
2 2
3
4
2) a - b = ( a - b )( a + a b + a b + ab + b ) ;
2
2
2
2
2
2
3) a 5 + b 5 = ( a + b )(a 4 - a 3b + a 2b 2 - ab 3 + b 4 ) ;
6
6
3
3
3
3
2
2
2
2
2
2
4) a -b = ( a -b )( a +b ) = ( a -b) ( a + ab +b ) ( a +b) ( a -ab +b ) = ( a -b ) ´
´ ( a 2 + ab + b 2 )( a 2 - ab + b 2 ) = ( a -b) ( a5 + a4b + a3b2 + a2b3 + ab4 + a5 ) ;
2
5) (a + b + c ) = a 2 + b 2 + c 2 + 2ab + 2ac + 2bc ;
2
6) (a - b - c ) = a 2 + b 2 + c 2 - 2ab - 2ac + 2bc ;
7) (a + b + c + d )2 = a2 + b2 + c2 + d 2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd ;
8) (a + b - c - d )2 = a2 + b2 + c2 + d 2 + 2ab - 2ac - 2ad - 2bc - 2bd + 2cd ;
9)
( am +b n ) ×( am -b n ) ×( am -bn ) ×( a m + ambn +b n ) = ( a m -b n ) ×( a m -b n ) ;
2
2
2
n
n
n-1
10) ( a + b) = a + na b +
2
2
4
3
3
n(n -1) n-2 2
n!
a b + ... +
an-kbk + ... + nabn-1 + bn ;
k !(n - k)!
2
11) ( a - b ) + ( b - c ) + ( c - a ) = 3 ( a - b )( b - c )( c - a ) .
3
3
3
Ba’zi yig’indilar
1. 1 + 2 + 3 + 4 +××× + n =
n(n + 1)
; 2. 2 + 4 + 6 + 8 +××× + 2n = n ( n +1) ;
2
3. 1 + 3 + 5 + 7 +×××+ 2n - 1 = n2 ;
4. 12 + 22 + ... + n2 =
æ 1ö æ 1ö
æ 1 ö n +1
5. ç1- 2 ÷ + ç1- 2 ÷ +... + ç1- 2 ÷ =
, (n ³ 2) ;
è 2 ø è 3 ø
è n ø 2n
13
n(n + 1)(2n +1)
;
2
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1
1
1
1
n
1 1 1
+
+
+ ... +
=
;
6. 1 + + + + ... = 2 ; 7.
1× 2 2 × 3 3× 4
n(n +1) n +1
2 4 8
n2 (n +1)2
1
1
1
1
n
9.
+
+
+ ... +
=
;
8. 1 + 2 + 3 + ... + n =
1× 3 3× 5 5× 7
(2n -1)(2n +1) 2n +1
4
3
3
3
3
10. 13 + 3 3 + 5 3 + 7 3 + ... + ( 2 n - 1 ) = n 2 ( 2 n 2 - 1) ;
3
n (4n 2 - 1)
11. 1 + 3 + 5 + 7 + ... + (2 n - 1) =
;
3
2
2
2
2
2
12. 1+ 2 + 22 +... + 2 n-1 = 2 n -1 13. 1× 2 + 2 × 5 + 3×8 +... + n ( 3n -1) = n2 ( n +1) ;
14. 1 - 2 2 + 32 - 4 2 + ... + ( -1) n -1 n 2 = ( -1) n -1
15. 1 × 2 + 2 × 3 + ... + ( n - 1) n =
n( n + 1)
;
2
( n - 1) n( n + 1)
.
3
Murakkab ildiz formulasi
a+b c =
1.
4.
k
k
x
x
n
k
y
x
k
k
x
x
n
k
a 2 - b 2c
2
y ... =
x ... =
k ×n
k
x:k x:
k
x:
k
a-
a 2 - b2c
.
2
xn × y .
1
x k -1
m ± m ± m ± m ±... =
6.
7.
a+
a-b c =
2.
3.
a + a 2 - b 2c
a - a 2 - b2c
+
.
2
2
n
. 5.
n -1
x x... x = 2 x2
±1+ 4m +1
, m> 0.
2
x ... =
k +1
x.
Ildizning xossalari
n
1.
a
(-a)2 = a .
n
ìï a ,
=í
ïî a ,
2.
n = 2k ,
k Î N,
n = 2 k + 1, k Î N .
a × b = a × b, a, b > 0.
14
, n = 1, 2,... .
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3. a b = a × b, a > 0.
2
2
4. a b = - a × b , a < 0.
a × 6 b5 = 6 a 2 × b5 .
a × 4 b = 4 a2 × b .
6.
7. n a = 0 , agar a = 0 bо`lsa. 8. n a = 1, agar a = 1 bо`lsa.
a
9. Agar a da a = 0, a > 0 bо`lsa, u hоlda a a =0 bо`ladi.
5.
3
a
a < 0 bо`lsa, u hоlda aa ratsiоnal
10. Agar a da a = 0,
kо`rsatkichli daraja ma`nоga ega emas.
11. a > 0, b > 0, c > 0; m , n , p Î N , m , n , p ³ 2 sonlar
uchun:
a)
n
a
m
=a
m
n
n
; b)
nm
n
n
n
g) a × b = a × b ; d)
j)
2n
a 2n = a ;
n× p
a× b;
n
n×k
e)
-a = -2n+1 a ;
a m× p = n a m ; o)
( )
l)
p
a
;
n
b
m
n
= am ;
n
a
n
a=
n×m
n×m× p
a n× pb p c ;
anb c =
m
n+k
n
v)
a = n×m a ;
2n+1
k)
2
m) a = a ; n)
a ×b =
n
n
a
=
b
am ;
n×k
k -n
a
; r) n a : k a = a
p) a × a =
.
12. Qisqa kо`рaytirish fоrmulalarni ildizli ifоdalarga qо`llanishi:
1) a - b = ( a - b )( a + b ), a ³ 0, b ³ 0;
2) a - b = ( - a - -b )( - a + -b ), a £ 0, b £ 0;
n
k
3) a - b =
(a - b)
4) a + b =
(a + b )
5)
n
a -
n
b =
(
2n
2
a - b ³ 0;
,
2
a + b ³ 0;
,
a -
2n
)(
6) a - b = ( 3 a -
3
b )( 3 a 2 +
3
7) a + b = ( a +
3
b)
3 - 2 2 = 12 - 2 2 ×1 +
8)
(
a2 -
3
( 2)
a +
2n
b
2
=
3
3
2n
ab +
ab +
(1 - 2 )
2
2
)
a ³ 0, b ³ 0;
b ,
3
3
b2 ) ;
)
b2 ;
= 2 - 1;
2
9) æçè 5 + 2 6 - 5 - 2 6 ö÷ø = æçè 5 + 2 2 3 - 5 - 2 2 3 ö÷ø =
æ
ç
è
(
3+ 2
)
2
-
(
3- 2
)
2
2
ö
÷ = 8.
ø
15
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Kasrning maxrajdagi irratsiоnallikdan(ildizdan)
qutqarish
5
5 × 3 a2 53 a2 53 a2
3 5 3 5 × 2 3 × 5 × 2 3 10
=
=
.
=
=
=
. 2. 3 =
1.
a
5×2
10
a 3 a × 3 a 2 3 a3
5 2 5 2× 2
(
)(
)
)
(
)
(
(
)(
)(
(1 - a )(1 + a ) 1 + a = (1 + a ) 1 + a .
1 - a 2 (1 - a 2 ) × 1 + a
=
=
3.
1- a
1 - a 1- a × 1 + a
(
4.
)
)
1+ 2 - 5
2 +1
1
1+ 2 - 5
1+ 2 - 5
=
=
=
=
2
1 + 2 + 5 1+ 2 + 5 1+ 2 - 5
2
2
1
2
1
+
1+ 2 - 5
(
)(
) (
)
(
)
2 + 2 - 10 + 1 + 2 - 5 3 + 2 2 - 10 - 5
.
=
2(2 - 1)
2
Chiziqli
tenglama
ax + b = 0 - chiziqli tenglama.
1. Agar a ¹ 0, b Î R bо`lsa, u hоlda ax + b = 0 tеnglama yagоna
x = - ba
уеchimga ega.
2. Agar a = 0, b ¹ 0 bо`lsa, u hоlda ax + b = 0 tеnglama
уеchimga ega emas, ya'ni уеchimlar tо`рlami Æ (bо`sh) bо`ladi.
3. Agar a = 0, b = 0 bо`lsa , u hоlda ax + b = 0 tеnglama
chеksiz kо`р уеchimga ega, ya'ni x Î R bо`ladi.
Kvadrat uchhadni chiziqli kо`рaytuvchilarga ajratish
2
1. ax + bx + c ( a ¹ 0) kо`rinishdagi ifоdaga kvadrat uchhad
dеyiladi, bu yеrda a , b , c Î R .
2. Agar D = b2 - 4ac > 0 bо`lsa, u hоlda kvadrat uchhadni quyidagicha
kо`рaytuvchilarga ajratamiz:
2
2
2
2
é
ù
2
éæ
ù
æ
ö
æ
ö
b
b
4
ac
b
D
æ
ö
ö
2
ax + bx + c = a êç x + ÷ - ç
÷ ú = a êç x + ÷ - çç
÷÷ ú =
÷ ú
êè
a
a
2a ø çè
2a
2
2
êè
ø è
ø úû
ø û
ë
ë
æ b D öæ b D ö æ -b + D öæ -b - D ö
= açç x + - ÷ç
÷ç x + 2a + 2a ÷÷ = açç x - 2a ÷ç
÷ç x - 2a ÷÷ = a(x - x1)× (x - x2),
2
a
2
a
è
øè
ø è
øè
ø
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bu yеrda
x1 =
-b+ D
,
2a
x2 =
-b- D
.
2a
2. Agar D = 0 bо`lsa, u hоlda kvadrat uchhad quyidagicha
kо`рaytuvchiga ajraladi:
2
b ö
æ
b
2
ax + bx + c = a ç x +
÷ = a ( x - x1 ) , bu yеrda x1 = x2 = - .
2a ø
2a
è
2
3. Agar D < 0 bо`lsa, u hоlda kvadrat uchhad chiziqli
kо`рaytuvchilarga ajralmaydi.
Kvadrat tеnglama va uning ildizlari
1. Kvadrat tenglamaning umumiy ko’rinishi ax2 + bx + c = 0 , a ¹ 0 ,
x –nо'malum. a, b, c –sоnlar kvadrat tеnglamaning kоeffitsiеntlari.
2. Kvadrat tenglamaning diskriminanti: D = b 2 - 4 ac .
2
3. Agar D º b - 4ac > 0 bо`lsa, u hоlda kvadrat tеnglama ikkita harxil haqiqiy ildizlariga ega bо`ladi: x1 =
-b+ D
-b - D
, x2 =
.
2a
2a
4. Agar D = 0 bо`lsa, u hоlda kvadrat tеnglama yagna haqiqiy ildizga
esa bо`ladi:
b
x1 = x2 = - .
2a
5. Agar D < 0 bо`lsa, u hоlda kvadrat tеnglama haqiqiy ildizlarga ega
bо`lmaydi, ya'ni Æ.
-b ± b2 - a × c
6. Agar ax + 2bx + c = 0 , a ¹ 0 bo’lsa, x1,2 =
bo’ladi.
a
2
2
p
æ pö
7. Agar x + px + q = 0 , D = ç ÷ -q bo’lsa, x1, 2 = - ±
2
è 2ø
2
D bo’ladi.
2
8. Agar a > 0, D > 0 bo’lsa, u hоlda ax + bx + c = 0 kvadrat tеnglama
uchun:
1) c > 0, b < 0 Þ x1 va x2 musbat yechimlar;
2) c > 0, b > 0 Þ x1 va x2 manfiy yechimlar;
3) c < 0 Þ x1 va x2 turli ishorali yechimlar.
2
9. x + px + q = 0 kvadrat tеnglama uchun:
2
1) q > 0, p > 0, p - 4 q ³ 0 Þ x1 va x 2 yechimlarga ega bo’ladi;
17
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p
2
p
q
4
³
0;
> C ; C 2 + pC + q > 0 Þ x1 va x2 ikkita
2)
2
yechimga bo`lib, x1 > C va x 2 > C bo’ladi, C - ixtiyoriy son;
2
3) p - 4 q ³ 0; -
p
< C ; C 2 + pC + q > 0 Þ
2
yechimga ega bo`lib,
x1 < C
x1 va x2 ikkita
va x2 < C bo’ladi;
2
4) C + pC + q < 0 Þ x1 va x2 ikkita yechimga ega bo`lib,
x1 > C va x2 < C bo’ladi.
Viet teoremasi
2
1. x1 va x2 sonlar ax + bx + c = 0 , a ¹ 0 tenglamaning ildizlari
ì x1 + x2 = - b a,
í
î x1 × x2 = c a .
bo’lsa:
2
2. x 1 va x2 sonlar x + px + q = 0 tenglamaning ildizlari bo’lsa:
ì x1 + x2 = - p,
í
î x1 × x2 = q.
Viet teoremasiga teskari teorema
ì x1 + x 2 = - b a ,
1. í x × x = c a
î 1 2
bo’lsa, x1 va x2 sonlar ax2 + bx + c = 0 , a ¹ 0
yoki a ( x - x1 )( x - x2 ) = 0 tenglamalarning ildizlari bo’ladi.
ì x1 + x 2 = - p
2. í x × x = q
î 1
2
bo’lsa, x1 va x2 sonlar x 2 + px + q = 0
yoki ( x - x1 )( x - x2 ) = 0 tenglamalarning ildizlari bo’ladi.
Kvadrat tеnglamaga kеltiriladigan tеnglamalar
1. ax
2n
+ bx n + c = 0 , a ¹ 0, n Î N , n ³ 2
Þ x = y Þ ay + by + c = 0 Þ y 1
n
2
Þ y1 = x n , y2 = x n
y1 = y2 = x n
agar
agar
2
- b ± b 2 - 4 ac
=
Þ
2a
b 2 - 4ac > 0;
b 2 - 4ac = 0; Æ agar b 2 - 4 ac < 0.
18
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2. Uchinchi darajali simmеtrik tеnglama:
x + 1 = 0,
é
ax 3 + bx 2 + bx + a = 0
Þ
ê ax 2 + ( b - a ) x + a = 0 .
ë
3. Tо`rtinchi darajali simmеtrik tеnglama:
1 ö æ
1ö
æ
ax 4 + bx 3 + cx 2 ± bx + a = 0 Û aç x 2 + 2 ÷ + bç x ± ÷ + c = 0
xø
x ø è
è
é a ( y 2 - 2) + by + c = 0,
1
® y= x±
® ê
2
x
ë a ( y + 2) + by + c = 0.
ö
2 æ
2
2
2
y
+
c
+
c
ax
+
bx
=
y
(
ax
+
bx
+
c
)(
ax
+
bx
+
c
)
=
d
Þ
4.
Þ
ç 1 ÷ y + cc1 - d = 0 .
1
è
ø
a +b
4
4
5. (x - a) + (x - b) = c Þ y = x almashtirish yordamida yechiladi.
2
4
2
6. Bikvadrat tеnglama: ax + bx + c = 0, a ¹ 0 Þ
x1, 2 = ±
·
·
·
-b +
b 2 - 4ac
;
2a
x3,4 = ±
-b -
b 2 - 4ac
.
2a
x1 + x2 + x3 + x4 = 0.
Ildizlari yi’indisi:
c
x
×
x
×
x
×
x
=
3
4
Ildizlari ko’paytmasi: 1 2
a .
Eng katta ildizining eng kichik ildiziga nisbati -1 ga teng.
Kvadrat tеnglama ildizlarini xossalari
0. x1 + x2 = - b a,
x1 × x2 = c a .
b 2 - 2 ac
1. x + x = ( x1 + x 2 ) - 2 x1 x 2 =
.
2
a
2
1
2
2
2
3
æ b ö 3bc
2. x + x = ( x1 + x2 )( x - x1x2 + x ) = ( x1 + x2 ) - 3x1x2 ( x1 + x2 ) = - ç ÷ + 2 .
èaø a
b
1 1 x1 + x2
1
1
x 12 + x 22
b 2 - 2ac
3.
+ =
= - . 4.
+ 2 =
=
.
x1 x2
x1 x2
c
x12
x2
x12 x 22
c2
3
1
5.
3
2
2
1
2
2
3
1
1
x13 + x 23
- b 3 + 3abc
.
+ 3 =
=
x13
x2
x13 x 23
c3
2
é b 2 - 2 ac ù
2c 2
6. x + x = éë ( x1 + x 2 ) - 2 x1 x 2 ùû - 2 x x = ê
ú - a2 .
2
a
ë
û
4
1
4
2
2
2
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2
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Kоmрlеks sоnlar. Kоmрlеks nоma'lumli kvadrat
tеnglamalar
Kоmрlеks sоn dеb z = a+bi kо`rinishidagi ifоdaga aytiladi, bunda a
va b lar haqiqiy sоnlar, i -shunday sоnki, i 2 = -1 , Re z = a, Im z = b.
1. Agar z 1 = a + bi vа z 2 = c + di
bо`lsa, U hоlda:
z1 = z 2
va
b = d bо`lsa
1) agar a = c
bо`ladi;
2) z1 + z 2 = (a + c ) + (b + d ) i ;
3) z1 - z 2 = (a - c ) + (b - d ) i ;
4) z1 × z2 = (ac - bd ) + (ad + bc ) i ;
z1
ac + bd
bc - ad
= 2
+ 2
i.
5)
2
z2
c +d
c + d2
| z |= a + b
2. Kоmрlеks sоnning mоduli
3. Kоmрlеks nоma`lumli kvadrat tеnglama:
az 2 + bz + c = 0
2
( a,
b, c ÎR, a ¹ 0, D = b2 - 4ac < 0) Þ z1,2 =
2
ga tеng .
-b + D
.
2a
Birinchi darajali ikki nоma'lumli ikkita
tеnglamalar sistеmasi
ìa1 x + b1 y = c1
í
tenglamalar sistemasi, bu уеrda a1, a2 , b1 , b2, c1, c2 îa2 x + b2 y = c2
2
2
2
2
bеrilgan sоnlar bо`lib, a1 + b1 ¹ 0 , a 2 + b2 ¹ 0 , x va y nоma'lum sоnlar.
a1
b1
¹
1. Agar a
bo’lsa, sistema yagona echimga ega.
b2
2
a1
b1
c1
=
¹
2. Agar a
bo’lsa, sistema echimga ega emas, ya'ni Æ.
b2
c2
2
a1
b1
c1
3. Agar a = b = c bo’lsa, sistema cheksiz ko’p echimga ega.
2
2
2
ìï a1 x + b1 y = c1
c1
b
= 1 bo`lganda yagona echimga ega.
4. í a x + b y = c sistema
c2
b2
ïî 2
2
2
ìï a1 x + b1 y = c1
c1 a1
í
5. a x + b y = c sistema c = a bo`lganda yagona echimga ega.
ïî 2
2
2
2
2
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Sistеmani yechish usullari
1. О`rniga qо`yish usuli:
1) Sistеmaning bir tеnglamasidan bir nоma`lumni ikkinchisi
оrqali ifоdalash; masalan, y ni x оrqali ifоdalash;
2) Hоsil qilingan ifоdani sistеmaning ikkinchi tеnglamasiga
qо`yish;
3) x ga nisbatan hоsil bо`lgan bir nоma`lum tеnglamani
yechish;
4) x ning tорilgan qiymatini y uchun ifоdaga qо`yib, y ning
qiymatini tорish kеrak.
2. Algеbraik qо`shish usuli:
1) Nоma`lumlardan birining оldida turgan kоeffitsiеntlar
mоdullarini tеnglashtirish;
2) Hоsil qilingan tеnglamalarni hadlab qо`shib yоki ayirib, bitta
nоma'lumni tорish;
3) Tорilgan qiymatni bеrilgan sistеmaning tеnglamalaridan biriga
qо`yib, ikkinchi nоma'lumni tорish kеrak.
Sonli oraliqlar
Kеsmalar, intеrvallar, yarim intеrvallar va nurlar sоnli
оraliqlar dеyiladi.
1. Ochiq oraliq(interval): a < x < b
x Î ( a, b )
2. Yopiq oraliq(kesma): a £ x £ b
x Î [a, b ] .
3. Yarim ochiq oraliq
(yarim interval):
a< x£b
x Î ( a, b ] ,
a£ x<b
a
b
x Î [a, b ) .
4. Nur(yarim tо`g`ri chiziq): a £ x < +¥
x Î [a, +¥) ,
-¥ < x £ a
x Î ( -¥, a ] .
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Tengsizliklar va ularning xossalari
1.
2.
3.
4.
5.
6.
7.
Agar
Agar
Agar
Agar
Agar
Agar
Agar
a > b bо`lsa, a - b > 0 bо`ladi.
a > b va b > c bо`lsa, a > c , a - c > 0 bо`ladi.
a > b bо`lsa, a ± c > b ± c bо`ladi.
a > b va c > 0 bо`lsa, a × c > b × c yоki a : c > b : c bо`ladi.
a > b va c < 0 bо`lsa, a × c < b × c yоki a : c < b : c bо`ladi.
a > b va c > d bo’lsa, a + c > b + d bо`ladi.
a > b va c < d bo’lsa, a - c > b - d bо`ladi.
8. Agar a > b > 0 bo’lsa,
1 1
< ,
a b
1 1
- <0
a b
bо`ladi.
n
n
9. Agar a > b > 0 bo’lsa, a > b (n Î N ) bо`ladi.
10. Agar a, b > 0 bo’lsa, a + b ³ 2 a × b bо`ladi.
11. Agar a > 0
1
a
+
³2
bo’lsa,
a
12. Agar a < 0
bo’lsa, a +
13. Agar ab >0
a b
+ ³ 2 bо`ladi.
bо`lsa,
b a
14. Agar ab <0 bо`lsa,
bо`ladi.
1
£ -2 bо`ladi.
a
a b
+ £ -2 bо`ladi.
b a
2ab
bо`ladi.
a+b
16. Agar a > 0 , b > 0 bо`lsa, a 3 + b 3 ³ a 2 b + ab 2 bо`ladi.
17. Agar a > 0 , b > 0 , c > 0 bо`lsa, a + b + c ³ ab + bc + ac bо`ladi.
18. Agar a > 0 , b > 0 , c > 0 bо`lsa, (a + b + c) 3 £ 9(a 3 + b 3 + c 3 ) bо`ladi.
15. Agar a > 0 , b > 0 bо`lsa,
ab ³
a3 + b3
a+b 3
³(
) bо`ladi.
19. Agar a > 0 , b > 0 bо`lsa,
2
2
20. Turli xil tеngsizliklar:
1
1
2
b
)
a
+
³ 2; v) a2 + b2 + c2 ³ ab + bc + ac ;
³
1;
2
2
a
a +1
2a
2
g) 2
£ 1;
d ) ( a + b ) ³ 4ab ; e) 8(a 4 + b4 ) ³ (a + b)4 ;
a +1
a) a 2 +
j ) a b + b c + ac £ 3 a bc ; a , b , c Î N ; h) (1 + a)n > 1+ an (a > 0) ;
k ) a 4 + b 4 + c 4 ³ abc ( a + b + c );
l ) 2a2 + b2 + c2 ³ 2a(b + c);
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m)
a£ x£ y£ z £t £b Þ
x z
a
+ ³2
.
y t
b
n
21. Agar n Î N
æ 1ö
2
£
bо`lsa,
ç1+ ÷ < 3 bо`ladi.
è nø
n
n
ænö
ænö
ç ÷ > 1 × 2 × 3 ... n > ç ÷ bо`ladi.
è2ø
è3ø
n +1
n £ n n! £
bо`ladi.
2
n
2
22. Agar n ³ 6
bо`lsa,
23. Agar n Î N
bо`lsa,
24. Agar n ³ 5
bо`lsa, 2 > n
25. Agar n Î N
n
bо`lsa, 2 > 2n + 1 bо`ladi.
26. Agar n > 0
n+2
bо`lsa,
( n + 1)
2
bо`ladi.
<
1
n bо`ladi.
Bir nоma'lumli tеngsizliklar va ularni уеchish
Ushbu f ( x) < g ( x), f ( x) > g ( x) , f ( x) £ g( x) va f ( x) ³ g ( x)
tеngsizliklarga bir nоma'lumli tеngsizliklar dеyiladi.
Shunday qilib, bir nоma'lumli tеngsizliklarni уеchish uchun:
1) Nоma'lum qatnashgan hadlarni chaр tоmоnga, nоma'lum
qatnashmagan hadlarni esa о`ng tоmоnga о`tkazish (1-xоssa);
2) О`xshash hadlarni ixchamlab, tеngsizlikni ikkala qismini
nоma'lum оldidagi kоeffitsiеntga (agar u nоlga tеng bо`lmasa)
bо`lish (2-xоssa) kеrak.
Tеng kuchli tеngsizliklar
Agar f 1 ( x ) < g 1 ( x ) va f 2 ( x ) < g 2 ( x ) tеngsizliklarning
yеchimlar tо`рlami aynan bir xil bо`lsa (yoki tеngsizliklar yеchimga
ega bо`lmasa), u hоlda ular tеng kuchli (ekvivalеnt) tеngsizliklar
dеyiladi, ya`ni f1 ( x) < g1 ( x) Û f 2 ( x) < g 2 ( x) .
Bir nоma'lumli chiziqli tеngsizliklar
Ushbu ax + b > 0, ax + b ³ 0, ax + b < 0 va ax + b £ 0
tеngsizliklarga bir nоma'lumli chiziqli tеngsizliklar dеyiladi, bunda
a ¹ 0, b Î R , x - nоma'lum.
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Noqat`iy:
{ ax + b £ 0 } :
ax + b ³ 0
é
ö
1) a > 0, b Î R bо`lsa, x Î ê - ; +¥ ÷
ì
bù ü
æ
í x Î ç -¥; - ú ý ;
aû þ
è
î
bù
æ
x
Î
-¥
;
ç
2) a < 0, b Î R bо`lsa,
a ûú
è
ì
é b
öü
í x Î ê - ; +¥ ÷ ý ;
ë a
øþ
î
b
a
ë
ø
3) a = 0, b > 0 bо`lsa, x Î (-¥; +¥)
4) a = 0, b < 0 bо`lsa,
x ÎÆ
5) a = 0, b = 0 bо`lsa, x Î (-¥; +¥)
{ x ÎÆ } ;
{ x Î (-¥; +¥)} ;
{ x Î (-¥; +¥)} .
Qat`iy:
ax + b > 0
{ ax + b < 0 } :
æ b
ö
x
Î
a
>
b
Î
R
0,
ç - ; +¥ ÷
1)
bо`lsa,
è a
ø
bö
æ
x
Î
-¥
;
ç
÷
2) a < 0, b Î R bо`lsa,
aø
è
3) a = 0, b > 0 bо`lsa, x Î (-¥; +¥)
xÎÆ
5) a = 0, b = 0 bо`lsa, x Î Æ
4) a = 0, b < 0 bо`lsa,
ì
bö ü
æ
í x Î ç -¥; - ÷ ý ;
aø þ
è
î
ì æ b
öü
í x Î ç - ; +¥ ÷ ý ;
øþ
î è a
{ x ÎÆ } ;
{ x Î (-¥; +¥)} ;
{ x ÎÆ } .
Bir nоma'lumli chiziqli tеngsizliklar sistеmasi
ìx > a
1. Agar a , b Î R ; a > b bо`lsa, í x > b
î
ìx < a
2. Agar a , b Î R ; a > b bо`lsa, í x < b
î
Û x > a Û x Î ( a ; +¥ ).
Û x < b Û x Î ( -¥ ; b ).
3. Agar a, b Î R; a > b bо`lsa,
ìx < a
Û b < x < a Û x Î (b; a ).
í
îx > b
4. Agar a , b Î R ; a > b bо`lsa,
ìx > a
í
îx < b
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ìx ³ a
Û x=a =b.
í
îx £ b
ì0 > a
Û x Î Æ.
í
x
b
yoki
x
b
<
>
(
)
î
5. Agar a , b Î R ; a = b bо`lsa,
6. Agar a , b Î R ; a > 0 bо`lsa,
Kvadrat tеngsizlik va uning yеchimi
ax 2 + bx + c > 0, ax 2 + bx + c < 0,
ax 2 + bx + c ³ 0, ax 2 + bx + c £ 0
kvadrat tеngsizliklar dеyiladi, bunda
kо`rinishdagi tеngsizliklar
x - nоma`lum,
a ¹ 0, b, c Î R.
Noqat`iy:
ax 2 + bx + c ³ 0
1) a > 0, D > 0,
{
ax 2 + bx + c £ 0
{ x Î [ x ; x ]} ;
x Î [ x1 ; x2 ] { x Î (-¥; x1 ] È [ x2 ; +¥)} ;
x1 < x2 bо`lsa, x Î (-¥; x1 ] È [ x2 ; +¥)
2) a < 0, D > 0, x1 < x2 bо`lsa,
3) a > 0, D < 0 bо`lsa, x Î (-¥; +¥)
4) a < 0, D < 0 bо`lsa, x Î Æ
5) a > 0, D = 0 bо`lsa, x Î (-¥; +¥)
6) a < 0,
}
1
2
{ x Î Æ };
{ x Î ( -¥ ; +¥ )} ;
{ x = x1 = x2 = - b 2a } ;
D = 0 bо`lsa, x = x1 = x2 = - b 2a { x Î ( -¥ ; +¥ )} .
Qat`iy:
ax 2 + bx + c > 0
{
ax 2 + bx + c < 0
}
1) a > 0, D > 0, x1 < x2 bо`lsa, x Î (-¥; x1 ) È ( x2 ; +¥)
{ x Î ( x ; x )} ;
1
2
2) a < 0, D > 0, x1 < x2 bо`lsa, x Î ( x1; x2 ) { x Î ( -¥; x1 ) È ( x2 ; +¥)} ;
3) a > 0,
D < 0 bо`lsa, x Î (-¥; +¥)
4) a < 0,
D < 0 bо`lsa, x Î Æ
5) a > 0,
D = 0 bо`lsa,
6) a < 0,
D = 0 bо`lsa, x ÎÆ
{ x ÎÆ } ;
{ x Î (-¥; +¥)} ;
x Î (-¥; x1 ) È ( x1 ; +¥) { x ÎÆ } ;
{ x Î (-¥; x1 ) È ( x1; +¥)} .
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Sоnlarning moduli
Sоnlarning mоdulini umumiy kо`rinishda quyidagicha yоzish
ì a, agar a ³ 0,
a
=
í
mumkin:
î - a , agar a < 0;
Masalan: -11 = -(-11) = 11, 2,5 = 2,5, 0 = 0.
Mоdulning xоssalari:
1.
2.
3.
4.
a ³ 0;
5. a = b Þ a = ± b ; 9.
2
2
10.
6. a = a ;
a ³ - a ; 7. a × b = a × b ; 11.
a
a
=
- a = a ; 8.
(b ¹ 0) ;
b
b
a ³ a;
a +b £ a + b ;
a - b ³ a - b
a - b £ a + b ;
é a > c,
a
c
c
(
0
)
>
>
Û
12.
ê a < - c; 13. a < c (c > 0) Û -c < a < c.
ë
x - a, agar x - a > 0 Þ x > a,
ì
ï
0,
agar x - a = 0 Þ x = a,
14. x - a = í
ï - ( x - a ), a g a r x - a < 0 Þ x < a.
î
Рarametrlarga bоg’liq bir nо`malumli tengsizliklarni
yechish
1. a( x - 4) > x - 5 Û ax - 4a > x - 5 Û ax - x > 4a - 5
Û (a - 1)x > 4a - 5 :
1) Agar a - 1 > 0
2) Agar a - 1 < 0
3) Agar a - 1 = 0
bо`ladi.
Û
Û
Û
a > 1 bо`lsa, u hоlda x >
a < 1 bо`lsa, u hоlda
x <
4a - 5
bо’ladi;
a -1
4a - 5
a -1
bо’ladi;
a = 1 bо`lsa, u hоlda 0 × x > -5 bо`lib, x Î R
Ratsiоnal tеngsizliklarni yеchish
Ratsiоnal tеngsizliklar quyidagicha yеchiladi:
1.
P( x)
P ( x)
> 0 Û P ( x ) Q ( x ) > 0. 2.
< 0 Û P ( x ) Q ( x ) < 0.
Q ( x)
Q ( x)
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3.
P ( x)
³0
Q ( x)
P( x)
ì P ( x )Q ( x ) ³ 0
£0
. 4.
Û í
(
)
Q
x
îQ ( x ) ¹ 0
ì P ( x )Q ( x ) £ 0
Û í
.
îQ ( x ) ¹ 0 .
Modulli tenglamalar
Moduli tenglamalar quyidagicha ekvivalent almashtirish bilan
yechiladi:
1. f ( x) = f ( x) Û f ( x) ³ 0 ; 2. f ( x) = - f ( x) Û f ( x) £ 0 ;
agar F ( x ) > 0,
é F ( x ) = f ( x ),
F
x
f
x
(
)
(
)
=
Û
3.
ê F ( x ) = - f ( x ), agar F ( x ) < 0;
ë
2
2
2
2
4. f ( x) = g ( x) Û f ( x) = g ( x) ; 5. f ( x) = a (a > 0) Û f ( x) = a ;
agar x - a ³ 0,
é F ( x , x - a ) = 0,
F
(
x
,
x
a
)
=
0
Û
6.
ê F ( x , - x + a ) = 0, agar x - a < 0;
ë
é f ( x) = g ( x),
=
Û
f
(
x
)
g
(
x
)
7.
ê f ( x) = - g( x);
ë
é f ( x ) = g ( x ), agar x ³ 0,
f
x
g
x
(
)
=
(
)
Û
8.
ê f ( - x ) = g ( x ), agar x < 0;
ë
é f ( x) = a
f
(
x
)
=
a
(
a
>
0)
Û
9.
; 10. f ( x) = a (a < 0) Û Æ .
ê
ë f ( x) = - a
Modulli tengsizliklar
Moduli tengsizliklar quyidagicha ekvivalent almashtirish bilan
yechiladi:
1. f (x) < a (a > 0) Û -a < f ( x) < a ;
2
2
2. f ( x) > a (a > 0) Û f ( x) > a yoki
é f ( x) > a,
f ( x) > a (a > 0) Û ê
agar a < 0 Þ x Î R;
ë f ( x) < -a;
2
2
3. f ( x ) < j ( x ) Û f ( x ) < j ( x ) ;
é f ( x ) < g ( x ), agar x ³ 0,
f
x
g
x
(
)
(
)
<
Û
4.
ê f ( - x ) < g ( x ), agar x < 0;
ë
ì f ( x) < g ( x),
f
(
x
)
<
g
(
x
)
Û
agar g ( x) £ 0 Þ x ÎÆ;
í
5.
î- f ( x) < g ( x);
27
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é f ( x ) > g ( x),
é f ( x) > g ( x ) agar x ³ 0,
f
x
g
x
yoki
(
)
>
(
)
Û
ê
ê
6.
êë f ( x ) < - g ( x)
êë f (- x) > g ( x ) agar x < 0;
2n
2
7. a f (x) +b f (x) +c ³0( £0) Þ f (x) =y Þay +by +c ³0( £0) ; y ³0, nÎN
n
n
Irrasional tenglama.
Irrasional
tenglamalarni
umumiy
holda
(n Î N ) :
ekvivalent almashtirish yordamida yechish mumkin
2n
ì f ( x) ³ 0,
ï
f ( x) = 2n j ( x) Û íj ( x) ³ 0,
2.
ï f ( x) = j ( x).
î
3.
2n
f (x) = a (a < 0) Þ xÎÆ.
4.
2n+1
1.
2n
quyidagicha
ì f ( x) ³ 0,
ï
f ( x) = j ( x) Û íj ( x) ³ 0,
ï
2n
î f ( x) = j ( x ).
2n+1
(x) .
f ( x) = 2n+1 j ( x) Û f ( x) = j ( x) . 5. 2n+1 f (x) = j(x) Û f (x) =j
ì f ( x ) ³ 0, ( a ³ 0), j ( x ) ³ 0,
ï
f ( x) - j ( x) = a Û í
2
f
x
a
x
(
)
(
)
.
=
+
j
ïî
ì f ( x ) ³ 0, j ( x ) ³ 0, b - j ( x ) ³ 0,
ï
f ( x ) + j ( x ) = b (b ³ 0) Û í
2
ïî f ( x ) = b - j ( x ) .
(
6.
)
(
7.
)
Irrasional tengsizliklar
Irrasional tengsizliklar quyidagicha ekvivalent almashtirish
yordamida yechiladi ( n Î N ) :
1.
3.
2n
2n+1 f (x) < g(x) Û f (x) < g2n+1(x).
2n
5.
ì f ( x ) ³ 0,
ï
f ( x ) < g ( x ) Û í g ( x ) > 0,
2.
ï f ( x ) < g 2 n ( x ).
î
éìg(x) < 0,
êí
êî f (x) ³ 0,
f (x) > g(x) Û ê
ìg(x) ³ 0,
êïí
êëïî f (x) > g2n (x).
4.
6.
28
2n
ì f ( x) ³ 0,
ï
f ( x) < 2 n g ( x ) Û í g ( x) ³ 0,
ï f ( x) < g ( x).
î
2n+1 f (x) < 2n+1 g(x) Û f (x) < g(x).
2n+1
f (x) > g(x) Û f (x) > g2n+1(x).
w.
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ì g ( x ) < 0,
ìï g ( x) > 0,
ï
f ( x)
U
>
Û
³
1
(
)
0,
f
x
í
í
2n
7. g ( x)
ïî f ( x) > g ( x).
ï
2n
î f ( x ) < g ( x)
2n f ( x)
ì g ( x ) < 0,
ïì g ( x ) > 0, f ( x ) ³ 0,
U
<
Û
1
í
í
8. g ( x )
2n
ïî f ( x ) < g ( x ).
î f ( x) ³ 0
2n
Arifmetik progressiya
1.
n- hadini topish formulasi: an = a1 + ( n - 1) d , n Î N ,
bu yerda
d - ayirmasi, a1 - birinchi hadi, an n-chi hadi, n - hadlari soni.
2. d - ayirmani toppish: d = a2 - a1 = a3 - a2 = a4 - a3 = ... = an - an -1
(
)
yoki d = a n - a m ( n - m ) .
3. Xossalari:
an -k + an + k
a k -1 + a k +1
=
a
a
=
a) k
yoki
n
2
2
tenglik bajarilsa { an } ketma-ketlik arifmetik progressiya bo’ladi;
b) an - am = ( n - m ) d ; an + am = ak + a p « n + m = k + p;
v) a1 + an = a2 + an -1 = a3 + an - 2 = ... = an - k + ak +1;
4. Dastlabki n ta hadi yig’indisi - S n :
1) S n = a1 + a2 + a3 + ... + an ; 2) S n - S n - 1 = a n ;
( a1 + a n ) n
2 a1 + d ( n - 1)
× n ; S n = n × a( n+1) 2 ;
2
2
m+n
S
=
S - Sn , m ¹ n ;
S
=
S
+
S
+
n
×
k
×
d
4) n+k
; 5) m + n
n
k
m-n m
3) S n =
;
Sn =
(
)
k
k
6) Sn = Sn + d × n × (k -1), S n - n dan k gacha bo`lgan sonlar yig;
7) a 2 + a 4 + ... + a 2 n = a1 + a 3 + ... + a 2 n - 1 + n × d ;
Geometrik progressiya
n -1
1. n- hadini topish formulasi: bn = b1q , n Î N , bu yerda
q -maxraji, b1 - birinchi hadi, bn n-chi hadi, n - hadlari soni.
29
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a) bn = b2 q
n-2
yoki bk = bk q
n-k
; bn + k = bn q , bn - k = bn q
k
-k
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;
n -1
= b2 q n - 2 = b3 q n - 3 = ... = bn - 2 q 2 = bn -1q ;
b) bn = b1 q
2. q -maxrajini toppish:
b
b
b
b b
b
b
b
q = 2 = 3 = ... = n ; q 2 = 3 = 4 ; q 3 = 4 = 5 ; q k = n ;
b1 b2
bn -1
b1 b2
b1 b2
bn -k
3. Xossalari:
2
3
a) bk = bk -1 × bk +1 ; bk = bk -1 × bk × bk +1 ;
b) bn × bm = bk × b p agar m + m = k + p ;
v) agar bk , bn , bm , b p ;
æb ö
k , n, p Î N bо`lsa, çç k ÷÷
è bn ø
k-p
æb
=ç k
ç bp
è
ö
÷
÷
ø
k -n
bо`ladi;
g) agar b1, b2 , b3 , ..., bn , musbat hadli geometrik progressiya uchun:
bn +1 = bn × bn + 2 = ... = b1 × bn +1 ;
bn = bn - k × bn + k .
4. Dastlabki n ta hadi yig’indisi - S n :
1) Sn = b1 + b2 + b3 + ... + bn ;
3) Sn =
b1(qn -1)
q -1
(
, Sn =
bnq - b1
q -1
)
2) S n - S n -1 = b n ;
, (q ¹ 1);
4) S n + m = 2 S n + S m ; 5)
2n
toq b1(q
Sn =
2
-1)
q -1
,S
2n
juft b2 (q
=
2
n
b1 + b 3 + . . . + b 2 n - 1
b2 + b 4 + ... + b 2 n
-1)
q -1
=
;
1
q ;
k
n -1 k
S n - k chi haddan boshlab n ta hadi yig’indisi;
6) S n = S n q
7) geometrik progressiya hadlari soni toq bo`lsa,
b2 + b4 + ... + b2 n = éë( b1 + b3 + ... + b2 n-1 + b2 n+1 ) - b2 n +1 ùû × q bо`ladi.
5. Agar geometrik progressiyada q < 1 , q ¹ 0 bo`lsa, bu
progressiya cheksiz kamayuvchi geometrik progressiya deyiladi.
S - cheksiz kamayuvchi geometrik progressiya hadlari yig’indisi:
b1
b1
b2
juft
, ( q < 1 ) ; S toq =
,
=
.
S =
S
1- q
1- q2
1- q2
6. Agar geometrik progressiyada q > 1 , bo`lsa, bu progressiya
o`suvchi geometrik progressiya deyiladi.
30
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Aralashmaga oid masalalar
Konsentrasiyasi x % , massasi M 1 bo’lgan eritma konsentrasiyasi y % , massasi M 2 bo’lgan eritma bilanaralashtirilsa, massasi
M × x + M2 × y
M 1 + M 2 konsentrasiyasi z % : z % = 1
bo’lgan
M1 + M 2
eritma hosil bo’ladi.
FUNKSIYA
Aniqlanish sohasi (an.s.)
j ( x)
bo’lsa, an.s. f ( x) ¹ 0 bo’ladi.
f ( x)
l.
y=
2.
y=
3.
y = 2n+1 f ( x) ,
2n
4. y = 2n
bo’lsa, an.s.
f ( x) ³ 0 bo’ladi.
nÎ N
bo’lsa, an.s.
-¥ < f (x) < ¥ bo’ladi.
nÎ N
bo’lsa, an.s.
f ( x ) > 0 bo’ladi.
f ( x) , nÎ N
1
,
f ( x)
5. y = log g ( x ) f ( x ) bo’lsa, an.s.
ì f ( x) > 0,
ï
í g ( x) > 0, bo’ladi.
ï g ( x) ¹ 1;
î
6. y = arccos f ( x); y = arcsin f ( x) bo’lsa, an.s. - 1 £ f ( x ) £ 1
bo’ladi.
p
7. y = tg f ( x )
bo’lsa, an.s. f (x) ¹ + p n, n ÎZ bo’ladi.
8. y = ctg f ( x)
9. y = arctg x
10. y = arcctg x
2
bo’lsa, an.s. f (x) ¹ p n, n ÎZ bo’ladi.
bo’lsa, an.s. x ÎR bo’ladi.
bo’lsa, an.s. x ÎR bo’ladi.
y ax 2 + bx + c;=
y x ;=
y ax; =
y sin x; =
y cos x bo’lsa,
11. =
an.s. x Î R bo’ladi.
k
12. y = , k Î R, k ¹ 0 bo’lsa, an.s. D( y) = ( -¥; 0) È( 0; +¥) bo’ladi.
x
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ì f ( x) ³ 0, g ( x) ³ 0,
ï
f ( x) ± g ( x)
j ( x) ³ 0, y ( x) ³ 0,
13. y =
bo’lsa, an.s. í
bo’ladi.
j ( x) ± y ( x)
ï
î j ( x ) ± y ( x) ¹ 0
Qiymatlar sohasi (q.s.)
x
1. y = a bo’lsa, q.s. E ( y) = ( 0; + ¥ ) bo’ladi.
2. y = log a f ( x), a > 0, a ¹ 1 bo’lsa, q.s. E ( y) = ( -¥; + ¥ ) bo’ladi.
2
2
2
2
3. y = asink x + bcos k x bo’lsa, q.s. E( y) = éë- a - b ; a + b ùû bo’ladi.
4. y = arccos x bo’lsa, q.s.
5. y = arcsin x bo’lsa, q.s.
E ( y ) = [ 0; p ] bo’ladi.
é p pù
E ( y ) = ê - ; ú bo’ladi.
2 2
û
æ p pö
bo’lsa, q.s. E ( y ) = ç - ; ÷ bo’ladi.
è 2 2ø
6. y = arctg x
ë
bo’lsa, q.s. E ( y ) = ( 0; p
7. y = arcctg x
x0 = -
bo’ladi.
( x0 , y 0 ) :
2
8. y = ax + bx + c parabolaning uchi
4 ac - b 2
,
y0 =
4a
)
b
2a
bo’lsa:
a) a > 0 bo’lsa, q.s. E ( y) = [ y0 ; +¥ ) bo’ladi;
b) a < 0 bo’lsa, q.s. E( y) = ( -¥; y0 ] bo’ladi.
2
9. y = ax + bx + c
(
funksiyda x 0 ,
a) a > 0 bo’lsa, q.s. E ( y ) = ëé
(
)
y ; +¥ )
b) a < 0 bo’lsa, q.s. E ( y ) = 0;
y 0 , y0 > 0 bo’lsa:
0
bo’ladi.
y0 ùû bo’ladi.
10. y = x bo’lsa, q.s. E ( y ) = [ 0; + ¥ ) bo’ladi.
k
11. y = , k Î R, k ¹ 0 bo’lsa, q.s. E ( y) = ( -¥; 0 ) È ( 0; + ¥ ) bo’ladi.
x
Funksiyaning juft va toqligi
1. f ( - x ) = f ( x )
2. f ( - x ) = - f ( x )
bo’lsa. funksiya juft.
bo’lsa, funksiya toq.
32
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3. Yuqoridagi ikkala tenglik ham bajarilmasa, funksiya juft ham,
toq ham emas.
2
4
2n
( n Î N ) , y = cos x - funksiyalar juft.
4. y = x , y = x , y = x
3
2 n -1
( n Î N ) , y = sin x , y = tgx,
5. y = x , y = x , y = x
y = ctgx - funksiyalar toq.
6. y = arcsin x , y = arctg x - funksiyalar toq.
2
7. y = x - 5 x + 2, y = x + 3 , y = arccos x , y = arcctg x funksiyalar juft ham toq ham emas.
8. Toq funksiyaning grafigi koordinatalar boshiga nisbatan simmetrik.
9. Juft funksiyaning grafigi OY o’qiga nisbatan simmetrik
10. Xossalari: a ) Juft ± Juft = Juft ; b ) Toq ± Toq = Toq ;
v ) J u ft ± T o q = J u ft h a m , to q h a m e m a s ;
g ) J × J = J ; J : J = J ; J ×T = T ; J : T = T .
d ) Juft + Son = Juft , Toq + Son = Juft ham, toq ham emas.
Davriyligi
Agar f (x +T) = f (x) bajarilsa, f (x) davriy funksiya bo’ladi. T -davr.
1. y = sinx , y = cosx funksiyalarning eng kichik musbat davri 2p .
2. y = tgx, y = ctgx funksiyalarning eng kichik musbat (e.k.m.)
davri p .
3. y = sinkx , y = coskx funksiyalarning e.k.m. davri T =
4. y = tgkx , y = ctgkx funksiyalarning e.k.m. davri T1 =
2p
.
k
p
.
k
m
m
5. y = sin (ax + b), y = cos (ax + b) funksiyalarning e.k.m. davri
2p
p
T
=
m - toq bo`lsa: T2 =
m - juft bo`lsa: 3
.
a teng;
a
m
m
6. y = tg (ax + b), y = c tg (ax + b) funksiyalarning e.k.m. davri T3 =
p
a.
7. Bir necha davriy funksiyalarning yig`indisidan iborat davriy
funksiyaning e.k.m. davrini toppish uchun qo`shiluvchi
funksiyalar e.k.m. davrlarining EKUK ini olish kerak.
Masalan: y = 7 c o s ( 2 x + 1) + 3 tg 0 , 5 x + 5 s in 4 x funksiyalarning
e.k.m. davrini toping: T1 =
2p
2p p
= p , T2 = 2p. T3 = =
2
4 2
33
EKUK æç p , 2p ,
è
pö
÷ = 2p .
2ø
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Chiziqli
funksiya
1. y = kx + b to’g’ri chiziq tenglamasi, bunda k = tga - to’g’ri
chiziqning burchak koeffisienti, α - funksiya grafigining OX
o’qining musbat yo’nalishi bilan tashkil qilgan burchagi.
2. y = kx + b funksiyaning grafigi OY o’qini ( 0; b ) nuqtada,
æ b
ö
OX o’qini ç - ; 0 ÷ nuqtada kesib o`tadi.
è k
ø
3. y = k1 x + b1 va y = k 2 x + b2 tenglama bilan berilgan to’g’ri
chiziqlar orasidagi j burchakni topish formulasi:
tgj =
k 2 - k1
, k1k 2 ¹ -1 .
1 + k1 × k 2
Xossalari:
a) k1 = k2 ikki to’g’ri chiziqning parallellik sharti;
b) k1 × k 2 = -1 ikki to’g’ri chizi qning perpendikulyarlik sharti;
v) k1 = k2 bo’lib, b1 = b2 da to’g’ri chizilar ustma-ust tushadi;
g) k1 = k 2 bo’lib, b1 = b2 da to’g’ri chizilar ustma-ust tushmaydi;
d) k1 ¹ k2 bo`lsa, to’g’ri chizilar kesishadi.
4. Ikki A( x1 , y1 ) va B ( x2 , y2 ) nuqtadan
o’tuvchi to’g’ri chiziq tenglamasi:
y - y1
x - x1
y1 - y2
=
k
=
y2 - y1 x2 - x1 ,
x1 - x2 .
5. M ( x0 , y0 ) nuqtadan o’tuvchi va burchak
koeffisienti k ga teng bo’lgan to’g’ri
chiziq tenglamasi:
y - y0 = k ( x - x0 )
6. Uchta A( x1 , y1 ) B ( x2 , y2 ) va C ( x3 , y3 ) nuqtaning bir to’g’ri
chiziqda yotish sharti:
y 3 - y1
x - x1
= 3
.
y 2 - y1
x 2 - x1
7. To’g’ri chiziqning umumiy ko’rinishdagi tenglamasi:
ax + by + c = 0 , a , b , c Î R .
8. M ( x0 , y0 ) nuqtadan ax + by + c = 0 to’g’ri chiziqqacha masofa:
34
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d = a x0 + b y 0 + c
a2 + b2 .
9. Parallel ax + by + c1 = 0 , ax + by + c2 = 0 to’g’ri chiziqlar
orasidagi masofa:
d = c 2 - c1
a
2
+ b2 .
10. a1 x + b1 y + c1 = 0 va a2 x + b2 y + c2 = 0 to’g’ri chiziqlar:
a1
b1
c1
=
¹
a) a
b2
c 2 bo’lsa, parallel bo’ladi;
2
a1
b1
c1
b) a = b = c
bo’lsa, ustma-ust tushadi;
2
2
2
a1
b1
v) a ¹ b bo’lsa, ular kesishadi.
2
2
11. To’g’ri chiziqning koordinata o’qlardan ajratgan kesmalarga
nisbatan tenglamasi:
x
y
+
= 1,
a
b
c =
a2 + b2
r
12. M ( x0 , y0 ) nuqtadan o`tib m = ( A; B ) vektorga perpendikulya
bo`lgan to’g’ri chiziqning tenglamasi: A( x - x0 ) + B ( y - y0 ) = 0 .
r
13. M ( x0 , y0 ) nuqtadan o`tib m = ( A; B ) vektorga parallel bo`lgan
x - x0
y - y0
=
to’g’ri chiziqning tenglamasi:
.
A
B
r
14. y = f ( x) funksiyani m = ( A ; B ) vektoriga parallel ko’chirsak
natijasida y - B = f ( x - A ) funksiya hosil bo’ladi.
15. y = kx + b to’g’ri chiziqqa y = a to’g’ri chiziqqa nisbatan
y = - kx + 2 a - b .
simmetrik to’g’ri chiziq
16. y = kx + b to’g’ri chiziqqa y = x to’g’ri chiziqqa nisbatan
1
b
y =
xsimmetrik to’g’ri chiziq
.
k
k
17. y = kx + b to’g’ri chiziqqa OY o’qiga nisbatan simmetrik
to’g’ri chiziq y = - k x + b .
18. y = kx + b to’g’ri chiziqqa OX o’qiga nisbatan simmetrik
to’g’ri chiziq y = - kx - b .
19. y = f ( x) funksiya grafigi x ® +¥ da y = kx + b og`ma
35
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f ( x)
, b=
asimtotaga ega bo`ladi, bu erda k= lim
x®+¥ x
20. Agar lim f ( x ) = ±¥ yoki
x® a +0
lim
x®+¥
[ f ( x) - kx] .
lim f ( x ) = ±¥ bo`lsa, u holda
x® a -0
x = a to`g`ri chiziq y = f ( x) funksiya grafigining vertical
asimtotagasi bo`ladi.
Kvadratik funksiya
2
1. y = ax + bx + c , a ¹ 0 kvadratik funksiyaning umumiy ko’rinishi.
2
2. y = ax + bx + c , a ¹ 0 kvadratik funksiyaning grafigi
paraboladan iborat:
a) a > 0 bo’lsa, parabola tarmoqlari yuqoriga yo’nalgan;
b) a < 0 bo’lsa, parabola tarmoqlari pastga yo’nalgan;
v) D > 0 bo’lsa, parabola OX o’qini ikkita nuqtada kesib o’tadi:
g) D = 0 bo’lsa, parabola OX o’qiga bitta nuqtada urinadi;
d) D < 0 bo’lsa, parabola OX o’qi bilan umuman kesishmaydi.
3. Parabola uchining koordinatalari topish A ( x0 , y0 ) :
b
,
x0 = 2a
4 ac - b 2
y0 =
.
4a
4. Parabolaning simmetriya o’qi:
x = x0 = -
b
.
2a
5. Aniqlanish sohasi: D ( y ) = ( -¥ ; +¥ ) .
6. Qiymatlar sohasi E ( y ) :
a) a > 0 bo’lsa, q.s. E( y) = [ y0 ; +¥ ) bo’ladi;
b) a < 0 bo’lsa, q.s. E( y) = ( -¥; y0 ] bo’ladi.
2
7. y = ax + bx + c parabola grafigi:
a) a > 0 parabola tarmoqlari yuqoriga yo’nalgan:
36
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b) a < 0 parabola tarmoqlari pastga yo’nalgan:
8.
y = ax 2 + bx + c parabolaning grafigining OX o’qi bilan
(
kesishish nuqtalari: x1 = -b - D
)
(
x2 = -b + D
2a
)
2a .
9. a > 0 bo’lsa, parabola x = x0 nuqtada minimumi y = y0 bo’ladi.
10. a < 0 bo’lsa, parabola x = x0 nuqtada maksimumi y = y0 bo’ladi.
a
Darajali funksiya y = x
l.
y = xn ,
nÎN :
D ( y ) = E ( y) = ( -¥; ¥ ) .
D( y ) = ( -¥; ¥ ) , E ( y) = [ 0; +¥ ) ,
2.
y = x-n = 1 xn ,
nÎ N :
D( y) = ( -¥;0) È ( 0; +¥) , E( y) = ( 0; +¥) ,
37
D ( y ) = E ( y ) = ( -¥;0 ) È ( 0; +¥ ) .
w.
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3. y = x ,
n
nÎ N :
D ( y ) = E ( y ) = ( -¥ ; ¥ )
D ( y ) = E ( y ) = [ 0; +¥ )
p q
4. y = x ,
p, q Î Z , q ¹ 0 :
D( y ) = E ( y ) = ( 0; ¥ ) .
D( y ) = E ( y ) = [ 0; +¥ ) ,
Grafiklarni o’zgartirish
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Funksiyaning o’sishi va kamayishi
1. Agar x1 , x2 Î ( a; b ) bo`lib x1 > x2 , f ( x1 ) > f ( x2 ) bo`lsa, u holda
y = f ( x) o’suvchi bo`ladi.
2. Agar x1 , x2 Î ( a ; b ) bo`lib x1 > x2 , f ( x1 ) < f ( x2 ) bo`lsa, u holda
y = f ( x) kamayuvchi bo`ladi.
Ko’rsatkichli funksiyaning xossalari va grafigi
x
Ko’rsatkichli funksiyaning ko’rinishi: y = a
( a > 0, a ¹ 1) .
1. Aniqlanish sohasi D( y ) = ( -¥; + ¥ ) barcha haqiqiy sonlar
39
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2.
3.
4.
5.
to’plami.
Qiymatlar sohasi E ( y) = ( 0; + ¥ ) barcha musbat haqiqiy sonlar
to’plami.
Ko’rsatkichli funksiya a > 1 bo’lganda barcha haqiqiy sonlar
to’plamida o’suvchi; agar 0 < a < 1 bo’lganda kamayuvchi.
Ko’rsatkichli funksiyaning grafigi (0; 1) nuqtadan o’tadi va OX
o’qidan yuqorida joylashgan.
Ko’rsatkichli funksiya juft ham, toq ham, davriy ham emas.
x
6. y = a funksiyaning grafigi:
D( y ) = ( -¥; + ¥ ) ,
E ( y) = ( 0; + ¥ ) .
Ko’rsatkichli tenglama
x
Ushbu a = b ( a > 0, a ¹ 1, b Î R ) ko`rinishdagi tenglamalarga
sodda ko’rsatkichli tenglama diyiladi. Bundan:
éagar a > 0, a ¹ 1, b £ 0 bo`lsa, teglama yechimga ega emas,
a) a = b Û ê
loga b
x
Û x = loga b;
êëagar a > 0, a ¹ 1, b > 0 bo`lsa, a = a
x
= 1 ( a > 0, a ¹ 1) Û f ( x) = 0.
b) a
Yechishda qo’llaniladigan asosiy ekvivalent almashtirishlar:
f ( x)
= aj ( x ) Û f ( x) = j ( x), (a > 0, a ¹ 1)
1. a
f ( x)
éagar f ( x) £ 0 bo`lsa, yechim yo ' q,
j ( x)
=
>
¹
Û
a
f
x
a
a
(
)
(
0,
1)
êagar f ( x) > 0 bo`lsa, j ( x) = log f ( x).
2.
ë
a
g ( x)
= f ( x ) quyidagi hollarda yechish mumkin:
3. f ( x )
a ) g ( x ) = 1; b ) f ( x ) = ±1; v ) g ( x ) > 0, f ( x ) = 0.
x
x
x
x
x
4. f (a ) = 0 (a > 0, a ¹ 1) Û t = a , f (t ) = 0 Û a = t1, a = t2 , ..., a = tk .
f ( x)
+ b × a f ( x ) + g × a f ( x ) =0 (a ¹ 0, b , g Î R; b 2 =ac ) Û
5. a × a
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æaö
Ûç ÷
èbø
f ( x)
æaö
= t, a t 2 + b t + g = 0 Û ç ÷
èbø
f (x)
æaö
= t1 , ç ÷
èbø
f ( x)
= t2 .
f ( x)
c 0 (a , b , g Î R ; a × =
b 1) Û
+ b × a f ( x) + =
6. a × a
)
)
)
Û a f ( x=
t , a t 2 + ct + b= 0 Û a f ( x=
t1 , a f ( x=
t2 .
7. 1 + 3 = 2 Û (1 2 ) + (
x 2
x
x
x
x
pö
æ pö æ
3 2 = 1 Û ç sin ÷ + ç cos ÷= 1 Û=
x 2.
6ø è
6ø
è
)
x
f ( x)
f ( x)
( a , b > 0; a, b ¹ 1) Û ( a b ) = 1 Û f ( x=) 0.
8. a = b
f ( x)
Ko’rsatkichli tengsizliklar
Ko’rsatkichli tengsizliklar ushbu ekvivalent almashtirish
yordamida yechiladi:
ì 0 < a < 1,
ì a > 1,
f ( x)
g ( x)
U
<
Û
a
a
í
í
1.
î f ( x ) > g ( x );
î f ( x ) < g ( x ).
2. [ f ( x ) ]
3. a
f (x)
g (x)
ì 0 < f ( x ) < 1,
>1Û í
î g ( x ) < 0;
U
ì f ( x ) > 1,
í
î g ( x ) > 0.
é f ( x ) > lo g a b , a > 1, b > 0 ,
ê
> b Û ê f ( x ) < lo g a b , 0 < a < 1, b > 0 ,
ê x Î D ( f ), a > 0 , b £ 0 .
ë
f ( x)
£b
4. a
( a > 0,
a ¹ 1, b £ 0 ) Û yechimga ega emas.
LOGARIFM
log a b = x Û a x = b , a ¹ 1, a > 0, b > 0 .
log a b
Bundan asosiy logarifmik ayniyatni a
a - logarifmning asosi har doim a ¹ 1, a > 0 .
Logarifmning xossalari
= b olamiz,
log a a = 1, a ¹ 1, a > 0 ;
1)
3)
2) log a 1 = 0 ;
loga ( X ×Y ) = loga X + loga Y , X > 0, Y > 0 ;
4)
lo g a b =
5)
æXö
loga ç ÷ = loga X - loga Y, X > 0, Y > 0 ; 6) log b p = p log b,
a
a
èY ø
1
; a , b > 0; a , b ¹ 1 ;
lo g b a
41
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7)
9)
p
log q b p = log a b, q ¹ 0,
a
q
log a b =
logc b
logc a
p, q Î R ;
, c ¹ 1, c > 0 ;
8) log a q
10)
a
1
p = log a b ;
q
log b c
log b
=c
log a
b
;
log a
a
= b b , loga b > 0 ;
11) loga b × logb c × ××× × logx y = loga y; 12) a
13) log e x = ln x -natural logarifm; 14) log10 x = lg x - o'nli logarifm;
15.
16.
17.
18.
19.
20.
21.
22.
23.
a > 1, 0 < b < 1 yoki 0 < a < 1, b > 1 bo `lsa, log a b < 0 bo `ladi;
a > 1, b > 1 yoki 0 < a < 1, 0 < b < 1 bo `lsa, log a b > 0 bo `ladi;
a > 1, b > c > 0 bo`lsa, log a b > log a c bo`ladi;
0 < a < 1, b > c > 0 bo`lsa,
log a b < log a c bo`ladi ;
0 < p < 1, a > b > 1 bo`lsa ,
log a p < log b p bo`ladi ;
p > 1, a > b > 1 bo`lsa,
log a p > logb p bo`ladi ;
p > 1, 0 < a < b < 1 bo`lsa,
log a p > log b p bo`ladi ;
0 < p < 1, 0 < a < b < 1 bo`lsa,
0 < p < 1, a > b > 0 bo`lsa,
log a p < log b p bo`ladi ;
log p a < log p b bo`ladi ;
24. p > 1, a > b > 0 bo`lsa, log p a > log p b bo`ladi .
Logarifmik funksiyalarning xossalari va grafigi
Logarifmik funksiyaning ko'rinishi: y = log a x, ( a > 0, a ¹ 1, x > 0 ) .
1. Aniqlanish sohasi: D( y ) = ( 0; + ¥ ) barcha musbat sonlar to'plami.
2. Qiymatlar sohasi: E ( y ) = ( -¥; + ¥ ) barcha haqiqiy sonlar to'plami.
3. Logarifmik funksiya aniqlanish sohasida agar a > 1 bo'lsa,
o'suvchi. Agar 0 < a < 1 bo'lganda kamayuvchi.
4. Agar a > 1 bo'lsa, logarifmik funksiya x > 1 da musbat
qiymatlar, 0 < x < 1 da esa manfiy qiymatlar qabul qiladi.
5. Agar 0 < a < 1 bo'lsa, logarifmik funksiya 0 < x < 1 da musbat
qiymatlar, x > 1 da esa manfiy qiymatlar qabul qiladi.
6. y = loga x logarifmik funksiya juft ham, toq ham, davriy ham emas.
7. Logarifmik funksiyaning grafigi (1; 0) nuqtadan o’tadi.
42
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8. y = log a x,
( a > 0, a ¹ 1, x > 0 )
D( y ) = ( 0; + ¥ ) ,
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funksiyaning grafigi:
E ( y ) = ( -¥; + ¥ ) .
Logarifmik tenglamalar
Ushbu log a x= b ( a > 0, a ¹ 1, b Î R ) ko`rinishdagi
tenglamalarga sodda logarifmik tenglama diyiladi.
Yechishda qo’llaniladigan asosiy ekvivalent almashtirishlar:
b
1. log a x = b Û x = a , x > 0 (a ¹ 1, a > 0) .
b
2. log a f ( x ) = b Û f ( x ) = a , f ( x ) > 0, b Î R ( a ¹ 1, a > 0) .
ìï f ( x ) > 0, j ( x ) > 0, j ( x ) ¹ 1,
3. log j ( x ) f ( x ) = b Û í
b
ïî f ( x ) = j ( x ).
ìï f ( x ) > 0 , a > 0 , a ¹ 1,
4. lo g a f ( x ) = j ( x ) Û í
j (x)
.
ïî f ( x ) = a
ì f ( x) > 0, g ( x ) > 0, a > 0, a ¹ 1,
5. log a f ( x ) = log a g ( x ) Û í f ( x) = g ( x ).
î
ì f ( x) > 0,
ìg(x) > 0,
ï
ï
log
log
(
)
1,
0,
=
Û
¹
>
A
A
f
x
A
yoki
í
íg(x) ¹ 1, A > 0,
(
)
(
)
f
x
g
x
6.
ï f ( x) = g( x);
ï f (x) = g( x).
î
î
ì f ( x ) > 0, g ( x ) > 0,
ï
og g ( x )
= al a
Û í a > 0, a ¹ 1,
(
)
f
x
7.
ï f ( x ) = g ( x ).
î
ìï f ( x ) > 0, g ( x ) > 0,
) log a m ( x ) ( a > 0, a ¹ 1) Û í
8. log a f ( x ) + log a g ( x=
) m ( x ).
ïî f ( x ) × g ( x=
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ìï g ( x) > 0,
9. ( 2n + 1) log a f ( x) =log a g ( x) ( a > 0, a ¹ 1, n Î N ) Û í 2 n+1
( x) =g ( x).
ïî f
ìï f ( x ) > 0,
2
log
(
)
log
(
)
0,
1,
=
>
¹
Î
Û
n
f
x
g
x
a
a
n
N
(
)
í 2n
10.
a
a
ïî f ( x)= g ( x).
11. f (log a x)= 0, a > 0, a ¹ 1 Û log a x= t , f (t )= 0.
12. log a x + log b x + log c x= d , a > 0, b > 0, c > 0, a ¹ 1, b ¹ 1, c ¹ 1, x > 0 Û
Û log a x +
log a x
log a b
+
log a x
= d.
log a c
Logarifmik tengsizliklar
Logarifmik tengsizliklar ushbu ekvivalent almashtirish
yordamida yechiladi:
1. lo g a
ì a > 1,
ì 0 < a < 1,
ï
ï
f ( x ) ³ b Û í f ( x ) > 0 , 2. lo g a f ( x ) ³ b Û í f ( x ) > 0,
ï f ( x) ³ ab.
ï
b
î
î f (x) £ a .
ì 0 < a < 1, g ( x ) > 0, ì a > 1, g ( x ) > 0,
ï
ï
U í f ( x ) > 0,
3. log a f ( x ) < log a g ( x ) Û í f ( x ) > 0,
ï f ( x ) > g ( x );
ï f ( x ) < g ( x ).
î
î
ì 0 < f ( x ) < 1,
ì f ( x ) > 1,
ïï
ïï
U
<
Û
>
g
x
a
g
x
lo
g
(
)
(
)
0,
í
í g ( x ) > 0,
4.
f ( x)
ï
ï
a
a
>
g
x
f
x
(
)
(
)
;
[
]
îï
îï g ( x ) < [ f ( x ) ] .
ì 0 < f ( x ) < 1,
ì f ( x ) > 1,
5. lo g f ( x ) g ( x ) > 0 Û í 0 < g ( x ) < 1 U í g ( x ) > 1 .
î
î
ì 0 < f ( x ) < 1,
ì f ( x ) > 1,
<
Û
U
g
(
x
)
0
í
í
f ( x)
î g ( x) > 1
î0 < g ( x) < 1.
ì 0 < f ( x ) < 1,
ì f ( x ) > 1,
lo
g
g
(
x
)
0
³
Û
U
í
í
7.
f (x)
g
x
0
<
(
)
£
1
î
î g ( x) ³ 1.
ì 0 < f ( x ) < 1,
ì f ( x ) > 1,
£
Û
g
x
U
lo
g
(
)
0
í
í
8.
f (x)
î g (x) ³ 1
î0 < g ( x) £ 1.
ì j ( x ) > 1,
ì f ( x ) > 0,
ï
ï
U í 0 < j ( x ) < 1,
9. lo g j ( x ) f ( x ) > lo g j ( x ) g ( x ) Û í g ( x ) > 0 ,
6. lo g
ï f ( x ) > g ( x );
î
44
ï f ( x ) < g ( x ).
î
w
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10.
lo g j ( x ) f ( x ) £ lo g j ( x )
ì j ( x ) > 1,
ï
g ( x) Û í f ( x) > 0,
U
ï f ( x ) £ g ( x );
î
ì g ( x) > 0,
ï
í 0 < j ( x ) < 1,
ï f ( x ) ³ g ( x ).
î
TRIGONOMETRIYA
Boshlang’ich tushunchalar
1. a
0
2. a rad
-gradusdan radianga o’tish: a r a d =
p
×a o .
o
180
180o
=
× a rad
p
-radiandan gradusga o’tish: a o
.
3. Ta`riflar:
1) sin a =
y
= y;
r
3) tg a =
y
,
x
5) tga =
sin a
cos a
2) cos a =
x
,
y
x ¹ 0 ; 4) ctg a =
6) ctga =
;
x
= x;
r
y ¹ 0;
cos a
.
sin a
Trigonometrik funksiyalar qiymatlari jadvali
Burchak α,
gradus(radian)
0° (0)
15° (π/12)
18° (π /10)
22,5° (π /8)
sin α
0
3 -1
2 2
Funksiyalar
cos α
tg α
1
0
3 +1
2- 3
ctg α
Mavjud emas
2+ 3
2 2
5 -1
5 -1
4
5+ 5
2 2
10 + 2 5
10 + 2 5
5 -1
2- 2 2
2+ 2 2
2 -1
2 +1
30° (π /6)
12
3 2
1
36° (π /5)
5- 5
2 2
5 +1
4
10 - 2 5
5 +1
10 - 2 5
45° (π /4)
60° (π /3)
2 2
2 2
1
1
90° (π /2)
1
3 2
12
0
45
3
3
Mavjud
emas
3
5 +1
1
3
0
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75° (5 π /12)
3 +1
2 2
180° (π)
270° (3 π /2)
0
-1
3 -1
2 2
-1
0
360° (2 π)
0
1
2+ 3
2- 3
0
Mavjud
emas
0
Mavjud emas
0
Mavjud emas
Trigonometrik funksiyalarning ishoralari
Asosiy trigonometrik ayniyatlar
1. cos 2 a + sin 2 a = 1 .
3. tg a × ctg a = 1 .
2
5. 1 + tg a =
1
.
cos 2 a
2. tga =
sin a
1
p
=
; a ¹ ( 2n + 1) , n Î Z .
cos a ctga
2
cos a
1
ctg
a
=
=
; a ¹ p n, n Î Z .
4.
sin a tga
1
2
+
ctg
=
a
1
; a ¹ p n, n Î Z .
6.
sin 2 a
Trigonometrik funksiyalarning birini ikkinchisi orqali
ifodalash
2
1. cos a = ± 1 - sin a = ±
2
2. sin a = ± 1 - cos a = ±
ctga
1 + ctg 2a
tg a
1 + tg 2a
1
1 - cos 2 a
=±
3. tg a = ctg a = ±
cos a
4. c tg a =
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1
= ±
tg a
1 - s in 2 a
= ±
s in a
46
1
=±
=±
1 + tg 2a
1
1 + ctg 2a
sin a
1 - sin 2 a
cos a
1 - cos 2 a
.
.
.
.
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a - ning qaysi chorakka tegishliligiga qarab "+" yoki "-" ishoradan
æ pö
a
Î
ç 0; ÷ , ya`ni I-chorakda bo`lsa,
biri olinadi. Masalan: Agar
è 2ø
æ 3p ö
1- formulada + olinadi; agar a Î ç p ; 2 ÷ , ya`ni III-chorakda
è
ø
bo`lsa, 1- formulada - olinadi.
Qo'shish formulalari
1. cos (a
2. cos (a
3. sin (a
4. sin (a
+ b)
- b)
+ b)
- b)
= cos a cos b - sin a sin b .
= cos a cos b + sin a sin b .
= sin a cos b + cos a sin b .
= sin a cos b - cos a sin b .
tg a ± tg b
ctga × ctg b m 1
5 . tg (a ± b ) =
6. ctg (a ± b ) =
.
.
ctga ± ctg b
1 m tg a × tg b
Karrali burchaklar
1. sin 2a = 2 sin a cos a .
2 tg a
3. sin 2a =
.
1 + tg 2a
2. cos 2a =
cos 2a - sin 2a
1 - tg 2a
4. cos 2a =
.
1 + tg 2a
2tga
2
2
6.
2
=
.
a
tg
5. cos 2a = 2cos a - 1 = 1 - 2 sin a .
2
1 - tg a
ctg 2a - 1
2
7. ctg 2a =
.
8. tg 2a =
.
2ctga
ctga - tga
9. sin 3a = 3sin a - 4 sin 3a .
10. cos 3a = 4cos 3a - 3cos a
3tga - tg 3a
11. tg 3a =
.
1 - 3tg 2a
3ctga - ctg 3a
12. ctg 3a =
.
1 - 3ctg 2a
14. cos 4a = 8cos 4a - 8cos 2a + 1.
13. sin 4a = cos a × ( 4sin a - 8sin3a ) .
15. tg 4a =
4tga × (1 - tg 2a )
1 - 6tg 2a + tg 4a
ctg 4a - 6ctg 2a + 1
16. ctg 4a =
.
4ctg 2a × ( ctg 2a - 1)
.
Darajasini pasaytirish
2
1. sin a =
1 - cos 2a
.
2
2
2. cos a =
47
1 + cos 2a
.
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3sin a - cos 3a
.
4
1 - cos 2a
2
tg
a
=
5.
1 + cos 2a .
3
3. sin a =
4
4
7. sin a - cos a = - cos 2a .
1
8
4
9. sin a = ( cos 4a - 4cos 2a + 3) .
3
4
1
4
4
4
11. cos a + sin a = + cos 4a .
4.
cos 3a =
3cos a + cos 3a
.
4
1 + cos 2a
2
ctg
a
=
6.
1 - cos 2a .
4
4
8. cos a - sin a = cos 2a .
1
( cos 4a + 4cos 2a + 3) .
8
5 3
6
6
12. cos a + sin a = + cos 4a .
8 8
4
10. cos a =
Yarim burchak uchun formulalar
a
a
+ sin 2 = 1 ;
2
2
1)
cos 2
3)
cos a = cos 2
5)
7)
a
a
cos ;
2
2
a
1 - cos a
sin
=
±
;
4)
2
2
2) sin a = 2 sin
a
a
- sin2 ;
2
2
1 + cos a
a
cos = ±
;
2
2
ctg
a
1 - cos a
tg
=
±
6)
2
1 + cos a ;
1 + cos a
2 a
cos
=
8)
;
2
2
1 - cos a
2 a
tg
=
10)
2 1 + cos a ;
a
1 + cos a
=±
2
1 - cos a ;
a 1 - cos a
=
9)
;
2
2
1 + cos a
2 a
a
a
ctg
=
tg 2 × ctg 2 = 1 ;
11)
;
12)
2 1 - cos a
2
2
sin a
a 1 + cos a
a 1 - cos a
sina
ctg
=
=
tg
=
=
13)
2
sin a
1 - cos a ; 14)
2
sin a
1 + cos a .
sin 2
Trigonometrik funksiyalarni yarim burchak tangensi
orqali ifodasi
1) sin a = 2 tg
a
2
a
3) tg a = 2 tg 2
æ
æ
2 a ö
2 a ö
æ
2 a ö
1
1
a
cos
=
tg
+
tg
+
tg
1
ç
÷ ç
÷;
ç
÷ ; 2)
2ø è
2ø
2ø
è
è
a
æ
æ
2 a ö
2 a ö
.
÷ 2 tg
ç 1 - tg
÷ ; 4) c tg a = ç 1 - tg
2ø
2 ø
2
è
è
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Ko'paytmani yig'indiga keltirish
1
é sin (a + b ) + sin (a - b ) ùû .
2ë
1
cos a × cos b = éë cos (a + b ) + cos (a - b ) ùû .
2
1
sin a × sin b = éë cos (a - b ) - cos (a + b ) ùû .
2
tg a + tg b
tg b - tg a
tg a × tg b =
=
.
ctg a + ctg b
ctg b - ctg a
ctga + ctg b ctg b - ctga
ctga × ctg b =
=
.
tga + tg b
tga - tg b
tga + ctg b tga - ctg b
tga × ctg b =
=
ctga + tg b tg b - ctga .
1. sin a × cos b =
2.
3.
4.
5.
6.
7. cos a × cos 2a × cos 4a × ... × cos 2 a =
n
sin 2 n +1 a
.
2 n +1 sin a
p
2p
3p
np
1
×
×
×
...
×
=
cos
cos
cos
cos
8.
.
2n + 1
2n + 1
2n + 1
2n + 1 2n
p
2p
1
a
)
cos
×
cos
=
, 5 = 2 × 2 + 1, n = 2 ;
Masalan:
5
5
22
2p
7p
1
p
b)
9.
cos
co s a × co s
15
× cos
15
× ... × cos
15
=
27
, 15 = 2 × 7 + 1,
n = 7.
sin 2a
a
a
a
× co s × ... × co s n =
a .
2
4
2
2 n + 1 sin n
2
Yig’indini ko’paytmaga keltirish
a - b a +b
a +b
a -b
sin
a
sin
b
=
2
sin
×cos
× cos
. 2.
2
2 .
2
2
a +b a -b
a +b
a -b
× sin
× cos
. 4. cosa -cos b =-2sin
.
3. cosa + cos b = 2cos
2
2
2
2
sin (a + b )
sin (a - b )
a
b
a
b
tg
+
tg
=
tg
tg
=
5.
6.
cos a × cos b .
cos a × cos b .
1. sina + sin b = 2sin
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sin (a + b )
7. ctga + ctg b = sin a × sin b .
sin ( b - a )
8. ctga - ctg b = sin a × sin b .
æ pö
sinx
+
cos
x
=
2
sin
çx+ ÷.
9.
è 4ø
2
10. 1 + cos a = 2cos
2
11. 1 - cos a = 2sin
a
.
2
æ pö
13. sin x - cos x = 2sin ç x - 4 ÷ .
è
ø
æ
15. 3 s in x + c o s x = 2 s in ç x +
è
a
;
2
pö
æ
sin
x
+
3
cos
x
=
2
sin
x
+
ç
÷.
12.
3ø
è
pö
æ
sin
x
3
cos
x
=
2
sin
x
ç
÷.
14.
3ø
è
p ö
÷.
6 ø
kx
( k + 1) x
x
× sin
sin .
2
2
2
kx
( k + 1) x
x
17. cos x + cos 2 x + cos 3 x + ... + cos kx = sin × cos
sin .
2
2
2
16. sin x + sin 2 x + sin 3 x + ... + sin kx = sin
18. sin a + sin 3a + sin 5a + ... + sin (2 n - 1)a = sin 2 n a sin a .
19. cos a + cos 3a + cos 5a + ... + cos (2n -1)a = sin na × cos n a sin a .
n+k
n - k +1
x
20. cos kx + cos (k + 1) x + ... + cos nx = cos
x × cos
x sin .
2
2
2
21. sin
a
3a
5a
(2n - 1)a
a
+ sin
+ sin
+ ... + sin
= (1 - cos na ) 2 sin .
2
2
2
2
2
Muhim trigonometrik shakl almashtirishlar
sin 3a
.
4
cos 3a
cos a × cos(60o - a ) × cos (60o + a ) =
.
4
tg 3a
tga × tg (60o - a ) × tg (60o + a ) =
.
4
ctg3a
ctga × ctg (60o - a ) × ctg (60o + a ) =
.
4
sin 8a
cos a × cos 2a × cos 4a =
.
8 sin a
sin 16a
cos a × cos 2a × cos 4a × cos 8a =
.
16 sin a
o
o
1. sin a × sin (60 - a ) × sin (60 + a ) =
2.
3.
4.
5.
6.
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7.
( sin x + cos x )
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2
= 1 + sin 2 x .
8. cos x - sin x = cos x - sin x = cos 2x .
4
4
2
2
1 + cos 2 2 x
sin 2 2 x 3 + cos 4 x
= 1=
9. cos a + sin a =
.
2
2
4
1
1
6
6
2
10. cos a + sin a = ( 5 + 3cos 4 x ) = 1 + 3cos 2 x .
8
4
1
8
8
11. cos a - sin a = cos 2 x ( 3 + cos 4 x ) .
4
4
4
(
)
Keltirish formulalari
γ
p
p
3p
3p
-a +a p -a p +a
-a +a 2p - a 2p +a
2
2
2
2
sin γ
cos γ
tg γ
ctg γ
cosα cosα sinα -sinα -cosα -cosα -sinα
sinα -sinα -cosα -cosα -sinα sinα cosα
ctgα -ctgα -tgα tgα
ctgα -ctgα -tgα
tgα -tgα -ctgα ctgα
tgα -tgα -ctgα
TRIGONOMETRIK
sinα
cosα
tgα
ctgα
FUNKSIYALAR
y = sinx funksiyaning xossalari va grafigi
1. Aniqlanish sohasi: barcha haqiqiy sonlar to'plami R = ( -¥; + ¥ ) .
2. Qiymatlar sohasi: E ( y ) = [ - 1;1] .
3. y = sinx funksiyaning eng kichik musbat davri T = 2p , ya'ni
sin( x + 2p ) = sinx, x Î R.
4. y = sinx funksiya toq, ya'ni sin ( - x ) = sinx.
é p
p
ù
ë2
2
û
5. Funksiya ê - + 2p n; + 2p n ú , n Î Z kesmalarda -1 dan 1
2
ë 2
û
gacha o'sadi.
3p
ép
ù
+ 2p nú , n Î Z kesmalarda 1 dan - 1
6. Funksiya ê + 2p n;
gacha kamayadi.
7. Funksiyaning nollari: sinx= 0 Û x= p n, n Î Z .
8. y= sinx funksiya x= p 2 + 2p n, n Î Z nuqtalarda eng katta
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qiymatga erishadi va u 1 ga teng.
9. y = sinx funksiya x = 3p 2 + 2p n, n Î Z nuqtalarda eng kichik
qiymatga erishadi va u - 1 ga teng.
10. Musbat qiymatlami: sinx > 0 Û x Î ( 2p n; p + 2p n ) , n Î Z .
11. Manfiy qiymatlami: sinx < 0 Û x Î (p + 2p n; 2p + 2p n ) , n Î Z .
12. y = sinx runksiyaning grafigi:
y = cosx funksiyaning xossalari va grafigi
1. Aniqlanish sohasi: barcha haqiqiy sonlar to'plami R = ( -¥; + ¥ ) .
2. Qiymatlar sohasi: E ( y ) = [ - 1;1] .
3. y = cosx funksiyaning eng kichik musbat davri T = 2p , ya'ni
cos ( x + 2p ) = cosx , x Î R.
4. y = cosx funksiya juft, ya'ni co s ( - x ) = cosx .
5. Funksiya [ -p + 2p n; 2p n ] , n Î Z kesmalarda - 1 dan 1
gacha o`sadi.
6. Funksiya [ 2p n ; p + 2p n ] , n Î Z kesmalarda 1 dan - 1
gacha kamayadi.
7. Funksiyaning nollari: cosx = 0 Û x = p 2 + p n, n Î Z .
8. y = cosx funksiya x = 2p n, n Î Z nuqtalarda eng katta
qiymatga erishadi va u 1 ga teng.
9. y = cosx funksiya x = p + 2p n, n Î Z nuqtalarda eng kichik
qiymatga erishadi va u -1 ga teng.
10. Musbat qiymatlami:
cosx > 0 Û x Î ( - p 2 + 2p n; p 2 + 2p n ) , n Î Z .
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11. Manfiy qiymatlami:
cosx < 0 Û x Î (p 2 + 2p n; 3p 2 + 2p n ) , n Î Z .
12. y = cosx funksiyaning grafigi:
y = tgx funksiyaning xossalari va grafigi
1. Aniqlanish sohasi: x ¹ p 2 + p n, n Î Z bo'lgan barcha haqiqiy
sonlar to'plami.
2. Qiymatlar sohasi: barcha haqiqiy sonlar to'plami R = ( -¥; + ¥ ) .
3. Funksiyaning eng kichik musbat davri
tg ( x + p ) = tgx, x Î D ( tg ) .
T = p , ya'ni
4. y = tgx funksiya toq, ya'ni tg ( - x ) = - tgx , x Î D ( tg ) .
5. Funksiyaning nollari: tgx = 0 Û x = p n, n Î Z .
6. Musbat qiymatlami: tgx > 0 Û x Î (p n ; p 2 + p n ) , n Î Z .
7. Manfiy qiymatlami: tgx < 0 Û x Î ( - p 2 + p n; p n ) , n Î Z .
8. y = tgx funksiya ( - p 2 + 2p n; p 2 + 2p n ) , nÎZ oraliqlarda o'sadi.
9. y = tgx funksiyaning grafigi:
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y = ctgx
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funksiyaning xossalari va grafigi
1. Aniqlanish sohasi: x ¹ p n, n Î Z bo'lgan barcha haqiqiy sonlar
to'plami.
2. Qiymatlar sohasi: barcha haqiqiy sonlar to'plarru R = ( -¥; + ¥ ) .
3. y = ctgx funksiyaning eng kichik mustbat davri T = p , ya'ni
ctg ( x + p ) = ctgx , x Î D ( ctg ) .
4. y = ctgx funksiya toq, ya'ni ctg (- x ) = -ctgx, x Î D ( ctg ) .
5. Funksiyaning nollari: ctgx = 0 Û x = p 2 + p n, n Î Z .
6. Musbat qiymatlami: ctgx > 0 Û x Î (p n; p 2 + p n ) , n Î Z .
7. Manfiy qiymatlami: ctgx < 0 Û x Î ( - p 2 + p n; p n ) , n Î Z .
8. y = ctgx funksiya (p n; p + p n ) , n Î Z oraliqlarda kamayadi.
9. y = ctgx funksiyaning grafigi:
TESKARI FUNKSIYANI TOPISH
y = f ( x ) funksiyaga teskari funksiyani topish uchun:
1) y = f (x) tenglamani x ga nisbatan yechiladi, ya`ni tenglikdan
x = g ( y ) hosil qilamiz;
2) hosil bo`lgan tenglikda x va y lar o'rni o`zaro almashtiriladi,
ya'ni x Û y va y = g ( x ) hosil bo'ladi;
3) funksiyaning aniqlanish sohasi hisobga olinadi.
Demak, y = g ( x ) funksiya berilgan f ( x ) ga teskari funksiya
bo'ladi. Masalan: y =
5
+ 4 ga teskari funksiyani toping.
x+2
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x ¹ -2 aniqlanish sohasi. 1) y - 4 =
2) x Û y Þ y =
Demak, y =
5
-2;
x-4
5
5
Þx+2 =
- 2;
x+2
y -4
3) D( y ) = ( -¥; 4 ) È ( 4; +¥ ) .
5
5
+ 4 ga teskari funksiya.
- 2 funksiya y =
x+2
x-4
TESKARI TRIGONOMETRIK
FUNKSIYALAR
ARKSINUS
1. y = arcsinx funksiya [ -1; 1] kesmada o'suvchi va bir qiyniatli
aniqlangan.
é p pù
2. Aniqlanish sohasi: D( y) = [ -1;1] . 3.Qiymatlar sohasi: E(y) = ê- ; ú .
ë 2 2û
4. Funksiya toq, ya'ni arcsin( - x ) = - arcsinx .
5. Arksinusning ba`zi qiymatlari:
x
0
arcsinx
0
1
2
p
6
2
2
p
4
3
2
p
3
1
p
2
1
2
p
6
-
2
2
p
4
3
2
p
3
-1
-
p
2
6. y = arcsinx funksiya grafigi:
a ) sin ( arcsinx ) = x , agar x Î [ -1;1] ;
b) arcsinx(sinx) = x,
c)
-
é p pù
agar x Î ê- ; ú ;
ë 2 2û
p
p
£ arcsinx £ .
2
2
ARKKOSINUS
1. y = arccosx funksiya [ -1; 1] kesmada kamayuvchi va bir qiymatli
aniqlangan.
2. Aniqlanish sohasi: D( y) = [ -1;1] . 3.Qiymatiar sohasi: E(y) =[ 0;p ] .
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4. Funksiya juft ham, toq ham emas.
5. arccos ( - x) = p - arccos x .
6. Arkkosinusning ba`zi qiymatlari:
x
0
arc cos x p
2
1
2
p
3
2
2
p
4
3
2
p
6
1
1
2
2p
3
-
0
-
2
2
3p
4
-
3
2
5p
6
-1
p
7. y = arccosx funksiya grafigi:
a ) cos ( arccosx ) = x , agar x Î [ -1;1] ;
b) arccos(cosx) = x,
agar x Î[ 0;p ] ;
c ) 0 £ arccosx £ p .
ARKTANGENS
1. y = arctgx funksiya ( -¥; +¥ ) oraliqda o'suvchi va bir qiymatli
aniqlangan.
2. Aniqlanish sohasi: D( y ) = ( -¥; +¥ ) .
3. Qiymatlar sohasi: E ( y ) = ( -0,5p ; 0,5p ) .
4. Funksiya toq, ya'ni arctg ( - x ) = - arctgx .
5. Arktangensning ba`zi qiymatlari:
x
0
arctgx
0
1
3
p
6
1
3
p
4
p
3
6. y = arctgx funksiya grafigi:
a ) tg ( arctgx ) = x , agar x Î ( -¥ ; +¥ ) ;
b)
c)
arctg (tgx) = x,
-
agar
p
p
< arctgx < .
2
2
æ p pö
x Îç - ; ÷;
è 2 2ø
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-
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ARKKOTANGENS
1. y = arcctgx funksiya ( -¥;+¥) oraliqda kamayuvchi va bir qiymaili
aniqlangan.
2. Aniqlanish sohasi: D( y) = ( -¥; +¥ ) .
3. Qiymatlar sohasi: E ( y ) = ( 0; p ) .
4. Funksiya juft ham, toq ham emas. arcctg (- x) = p - arcctgx .
5. Arkkotangensning ba`zi qiymatlari:
0
1
-1 - 3
1
3 - 1
3
3
arcctgx p
p
p
p
2p
3p
5p
2
3
4
6
3
4
6
6. y = arctgx funksiya grafigi:
a ) c tg ( arcctgx ) = x , agar x Î ( -¥ ; +¥ ) ;
agar x Î( 0;p ) ;
b) arcctg(ctgx) = x,
c ) 0 < arcctgx < p .
Teskari trigonometrik funksiyalar ustida amallar
p
.
2
1. arcsin x + arccos x =
p
2
2. arctgx + arcctgx = .
2
3. sin(arccos x) = ± 1 - x ,
5. tg ( a rcctgx ) =
7.
tg ( arcsin x ) =
9. sin ( arctg x ) =
11. cos ( arctg x ) =
1
,
x
x £ 1. 4. cos (arcsin x) = ± 1 - x 2 , x £ 1.
1
ctg
arctgx
=
, x ¹ 0.
x ¹ 0.
(
)
6.
x
x
2
x < 1. 8. tg ( arccos x ) = ± 1 - x ,
,
± 1 - x2
x
± 1+ x
1
2
± 1+ x
2
x < 1.
x
.
10. sin ( arcctg x ) =
.
12. cos ( arcctg x ) =
(
(
1
± 1+ x
x
± 1+ x
)
)
.
2
2
.
ì - arccos xy + 1 - x 2 × 1 - y 2 , x > y ,
ï
13. arccos x - arccos y = í
ï arccos xy + 1 - x 2 × 1 - y 2 , x < y .
î
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x+ y
x- y
, xy < 1. 15. arctgx - arctgy = arctg
, xy > 1.
14. arctgx + arctgy = arctg
1 - xy
1 + xy
xy +1
xy -1
, x ¹ y.
, x ¹ - y. 17. arcctgx - arcctgy = arcctg
16. arcctgx + arcctgy = arcctg
x- y
x+ y
x
1- x2
,
, 0 < x £ 1. 19. ctg (arccos x ) =
18. ctg (arcsin x) =
x
1- x2
x < 1.
20. sin(2arcsin x) = 2 x 1 - x 2 ,
x £ 1.
21.
x £ 1.
2
22. cos(2arccos x) = 2x -1,
x £ 1.
2
22. cos(2arc s in x) = 1 - 2 x ,
23. tg (2arctg x) =
25.
2x
, x ¹ 1.
1- x2
sin (2 arccos x ) = 2 x 1 - x 2 ,
24. sin(2arctg x) =
1 - x2
, - ¥ < x < +¥.
cos(2arctg x) =
1 + x2
2x
, -¥ < x < +¥.
1+ x2
26. sin(2arcctg x) =
2
2
27. cos(2arcctg x) = - (1 - x ) (1 + x ) , - ¥ < x < +¥.
2x
, - ¥ < x < +¥.
1 + x2
Trigonometrik tenglamalar
1. sinx = a , a £ 1 Û x = ( -1) arcsina + p n , n Î Z .
Xususiy hollar:
a) sinx = 0 Û x = p n, n Î Z ;
b) sinx = 1,
Û x = p 2 + 2p n,
n Î Z;
v) sinx = -1, Û x = - p 2 + 2p n, n Î Z ;
n
g ) sin2 x = a , 0 £ a £ 1 Û x = ± arcsin a + p n, n Î Z .
2. cosx = a , a £ 1 Û x = ± arccosa + 2p n , n Î Z .
Xususiy hollar:
a) cosx = 0
Û x = p 2 + p n, n Î Z ;
b) cosx = 1,
Û x = 2p n,
nÎ Z;
v) cosx = -1, Û x = p + 2p n, n Î Z ;
g ) cos 2 x = a, 0 £ a £ 1 Û x = ± arccos a + p n, n Î Z .
3. tgx = a , a Î R Û x = arctga + p n , n Î Z .
Xususiy hollar:
a) tgx = 0 Û x = p n, n Î Z ;
b ) tgx = ± 1,
Û x = ± p 4 + p n,
nÎ Z;
v) tg 2 x = a, 0 £ a < +¥ Û x = ±arctg a + p n, n Î Z .
4. ctgx = a , a Î R Û x = arcctga + p n , n Î Z .
58
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Xususiy hollar:
a ) ctgx = 0 Û x = p 2 + p n, n Î Z ;
b ) c t g x = ± 1,
x = ± p 4 + p n,
Û
nÎ Z;
v) ctg 2 x = a, 0 £ a < +¥ Û x = ± arcctg a + p n, n Î Z .
a
b
c
a
×
sinx
+
bcosx
=
c
Û
sinx
+
cosx
=
Û
5.
2
2
2
2
2
2
a +b
a +b
a +b
c
Ûsinx×cosj +cosx× sinj =
bunda
cosj = a
a +b
2
a 2 + b2 ,
2
Ûsin(x +j) =
sinj = b
c
a +b
2
a2 + b2 ,
2
,
c
a +b
2
2
£1,
tgj = b a .
é ax + b - ( cx + d ) =2p n,
sin
(
ax
+
b
)
=
sin
(
cx
+
d
)
Û
ê
6.
êë ax + b + cx + d =( 2n + 1) p , n Î Z .
é ax + b - ( cx + d ) = 2p n,
cos
(
ax
b
)
cos
(
cx
d
)
+
=
+
Û
ê
7.
ë ax + b + cx + d = 2p n, n Î Z .
ì ax + b m ( cx + d ) = p n, n Î Z ,
ï
tg (ax + b ) = ±tg (cx + d ) Û í
p
p
8.
+
¹
+
+
¹
+ p n.
ax
b
n
,
cx
d
p
ï
î
2
2
ìïax + b m ( cx + d ) = p n,
+
=
±
+
Û
ctg
(
ax
b
)
ctg
(
cx
d
)
í
9.
ïîax + b ¹ p n, cx + d ¹ p n, n Î Z .
Trigonometrik tengsizliklar
1. sinx > a, a £ 1 Û x Î ( arcsina + 2p n; - arcsina + 2p n ) , n Î Z .
2. sinx ³ a , a £ 1 Û x Î [ arcsina + 2p n; - arcsina + 2p n ] , n Î Z .
3. sinx £ a , a £ 1 Û x Î [p - arcsina + 2p n; arcsina + 2p n ] , n Î Z .
4. cosx ³ a, a £ 1 Û x Î [ -arccosa + 2p n; arccosa + 2p n ] , n Î Z .
5. cosx £ a, a £ 1 Û x Î[ arccosa + 2p n; - arccosa + 2p (n + 1] , n Î Z .
6. tgx ³ a , a Î R Û
7. tgx £ a, a Î R Û
8. ctgx £ a , a Î R
9. ctgx ³ a , a Î R
Û
Û
x Î [ arctga + p n; p 2 + p n ) , n Î Z .
x Î ( - p 2 + p n; arctga + p n ] , n Î Z .
x Î [ arcctga + p n; p + p n ) , n Î Z .
x Î (p n; arcctga + p n ] , n Î Z .
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10 arctgx > arctgy Û x > y.
11. arcctgx > arcctgy Û x < y.
12. arcsinx > arcsiny Û - 1 £ y < x £ 1.
13. arccosx > arccosy Û - 1 £ x < y £ 1.
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Kvadratik, ko`rsatkchli, logarifmik, trigonomеtrik
funktsiyalari o`zining aniqlanish sohasida uzluksiz.
FUNKSIYANING LIMITI
Agar ixtiyoriy e > 0 son uchun shunday d > 0 son topilsaki,
argument x ning 0 < x - a < d tengsizlikni qanoatlantiruvchi barcha
qiymatlarida f ( x) - b < e tengsizlik bajarilsa, b son f ( x)
nuqtadagi ( x ® a dagi ) limiti deb ataladi
quyidagicha yoziladi: lim f ( x ) = b.
funksiyaning
a
va
x®a
1. Limitning xossalari: Agar lim f ( x) = A
lim g ( x ) = B
va
x® a
x® a
limitlar mavjud bo`lsa, u holda:
a)
lim
x® a
[ f ( x) ±
g ( x ) ] = lim f ( x ) ± lim g ( x ) = A ± B ;
x® a
x® a
lim [ f ( x ) × g ( x ) ] = lim f ( x ) × lim g ( x) = A × B;
b)
x ®a
v)
g)
lim
x® a
[ f (x)
x ®a
x®a
g ( x ) ] = lim f ( x )
lim g ( x ) = A B , B ¹ 0;
x® a
x® a
lim [ C × g ( x ) ] = C × lim g ( x ) = C × B
x®a
x®a
2. Ajoyib limitlar:
sin x
x
= lim
= 1.
1. lim
x ®0 x
x ®0 sin x
bo`ladi.
n
æ 1ö
6. lim ç1 + ÷ = e = 2,71183... .
n®¥ è
nø
1
sin px
px
= lim
= p, p Î R . 7. lim (1 + x) x = e .
2. lim
x®0
x®0 sin x
x®0
x
3. lim
tg x
x
= lim
= 1.
x® 0 x
x®0 tg x
8.
lim x x = 1 .
x ®0
ax -1
= ln a, a > 0 .
4. lim
x®0 x
9.
arcsin x
x
= lim
= 1.
x®0
x®0 arcsin x
x
ln ( x + 1)
= 1.
5. lim
x®0
x
10.
60
lim
a
x + 1) - 1
(
lim
=a,
x®0
x
a ¹ 0.
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HOSILA
1. x va
x0 - erkli o`zgaruvchilar y = f ( x) funksiyaning aniqlanish
sohasidan olingan qiymatlar bo`lsin, Dx = x - x0 ayirma erkli
o`zgaruvchining x0 nuqtadagi orttirmasi deyiladi.
Bundan x = x0 + Dx .
2. Dy º Df ( x0 ) = f ( x0 + Dx) - f ( x0 ) ga y = f ( x) funksiyaning x0
nuqtadagi orttirmasi deyiladi. Bundan f ( x0 + Dx) = f ( x0 ) + Df ( x0 ) .
3. y = f ( x) funksiyaning x 0 nuqtadagi hosilasi:
D f ( x0 )
f ( x0 + D x ) - f ( x0 )
y ¢ = lim
= lim
= f ¢( x0 ).
Dx ® 0
Dx ® 0
Dx
Dx
4. Hosilaning fizik va mexanik ma`nosi. Moddiy nuqta S = S ( t )
qonuniyat bilan harakatlanayotgan bo`lsa, u holda:
a) S ¢(t ) = J (t ) - harakat tezligi; b) S ¢¢ ( t ) = a ( t ) - harakat
tezlanishi bo`ladi.
5. Hosilaning giometrik ma`nosi. y = f ( x) funksiya grafigiga x0
nuqtada o`tqazilgan urinmaning burchak koeffisienti k va OX
o`qining musbat yo`nalishi bilan xosil qilgan burchagi a bo`lsa, u
holda: a) k = f ¢( x0 ); b) tga = f ¢( x0 );
v) y = f ( x) funksiyaga x = x0
nuqtada o`tqazilgan urinma tenglamasi:
=
y
6.
f ( x0 ) + f ¢( x0 ) ( x - x0 ) .
( y - y0 ) f ¢( x0 ) + ( x - x0 =)
0
- normal tenglamasi.
7. y = f (x) va y = g(x) funksiyalarga x = x0 nuqtada o`tqazilgan
urinmalar uchun:
a) f ¢( x0 ) = g ¢( x0 ) - parallellik sharti;
b) f ¢( x0 ) × g ¢( x0 ) = -1 - perpendikulyarlik sharti.
8. y = f ( x) va y = g ( x) funksiyalarga M ( x0 , y0 ) nuqtada
o`tqazilgan urinmalar orasidagi burchakni topish:
a)
tgj =
g ¢( x0 ) - f ¢( x0 )
, agar
1 + f ¢( x0 ) × g ¢( x0 )
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1 + f ¢( x0 ) × g ¢( x0 ) ¹ 0;
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b)
j = 90 ,
0
agar
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1 + f ¢( x0 ) × g ¢( x0 ) = 0.
y = f ( x) funksiya grafigiga tegishli bo`lmagan M ( x1 , y1 )
nuqtadan o`tib y = f ( x) funksiyaga uringan urinmaning
9.
urinish nuqtasini topish formulasi:
ìï y1 - y0 = f ¢( x0 ) ( x1 - x0 ) ,
í
ïî f ( x0 ) = y0 .
10. Agar f ¢¢( x) = 0 bo`lsa, x = xi , i = 1, 2,... nuqtalar y = f ( x )
funksiyaning egilish nuqtalari bo`ladi.
11. Agar f ¢¢( x) £ 0 [ f ¢¢( x) ³ 0] bo`lsa, u holda y = f ( x)
funksiyaning grafigi
( a, b ) intervalda
qavariq [botiq] bo`ladi.
Sodda funksiyalarning hosilasi
( C )¢ = 0,
1.
C = const. 2. ( x )¢ =1. 3.
1
æ 1 ö¢
=
.
ç ÷
x2
èxø
5.
9. loga x ¢ =
(
)
6.
( e x )¢ = e x .
7.
( xa )¢ = a xa -1.
4.
( a x )¢ = a xln a.
( x )¢ = 2 1 x .
8.
( ln x )¢ =
1
.
x
1
1
. 10. ( sin x )¢ = cos x. 11. ( cos x )¢ = -sin x. 12. ( tg x )¢ = 2 .
xlna
cos x
13. ( ctg x)¢ = -
1
1
1
¢
¢
arcsin
x
arccos
x
.
14.
=
.
15.
=
.
(
)
(
)
2
2
sin2 x
1- x
1- x
1
1
¢
1 6 . ( a r c tg x )¢ =
.
1
7
.
a
r
c
c
tg
x
=
.
(
)
1 + x2
1 + x2
Hosilalarni hisoblash qoidalari
Agar u = u ( x) va J = J ( x ) bo'lsa, u holda:
1) ayirma va yig'indining hosilasi: ( u ± J )¢ = u ¢ ± J ¢;
× u )¢ = c × u ¢
2) agar c = const bo'lsa,
(c
3) ko'paytmaning hosilasi:
( u ×J )¢ = u ¢ × J + u ×J ¢ ;
æu
4) bo'linmaning hosilasi: ç J
è
;
ö¢ u ¢ × J - u × J ¢
÷ =
.
J2
ø
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Murakkab funksiyaning hosilasi
1.
(
f (x)
)
¢
=
f ¢( x )
.
2
f (x)
2.
æ
ö¢
1
f ¢( x )
.
ç
÷ = 2
f
(
x
)
f
(
x
)
è
ø
f ¢( x ).
4.
( a )¢ = a
3.
( e )¢ = e
5.
f ¢( x)
¢
=
lnf
x
(
)
.
(
)
f ( x)
6.
f ¢( x)
¢
log
f
x
(
)
=
.
( a
)
f ( x)lna
7.
( sinf ( x))¢ = cos f ( x) × f ¢( x).
8.
( cos f ( x) )¢ = -sin f (x) × f ¢( x).
f ( x)
f ( x)
f ¢( x )
.
cos 2 f ( x )
f ¢( x )
11. ( arcsinf ( x ) )¢ =
.
1 - f 2 ( x)
( tg
9.
f ( x) )¢ =
13. ( arctg f ( x) )¢ =
(
15.
17.
f ¢( x)
.
1+ f 2 ( x)
n
× lna × f ¢( x).
f ( x) )¢ = -
f ¢( x)
14. ( arcctg f ( x) )¢ = .
1+ f 2 ( x)
f ¢(x)
¢
16. n f ( x) =
.
n n-1
n × f ( x)
)
1
f ( x)
f ( x)
f ¢( x )
.
sin 2 f ( x )
f ¢( x )
12. ( arccosf ( x ) )¢ = .
1 - f 2 ( x)
( ctg
10.
¢
f a ( x) = a f a -1(x) f ¢( x).
æ
çç
è
f ( x)
(
)
ö¢
f ¢( x )
.
÷÷ = n +1
n
n× f
(x)
ø
Funksiyaning o'sish va kamayish oraliqlari
1. Agar y = f ( x) funksiya ( a, b ) intervalda differensiallanuvchi va
f ¢( x ) > 0, bo`lsa, u holda y = f ( x ) funksiya shu intervalda
o`sadi.
2. Agar y = f ( x ) funksiya ( a, b ) intervalda differensiallanuvchi va
f ¢( x ) < 0, bo`lsa, u holda y = f ( x) funksiya shu intervalda
kamayadi.
3. Agar y = f ( x) funksiya yopiq [ a, b ] oraliqda uzliksiz boqlib,
( a,b) intervalda
differensiallanuvchi va f ¢( x ) > 0 ( f ¢( x ) < 0 ) ,
bo`lsa, u holda y = f ( x ) funksiya yopiq [ a, b ] oraliqda
o`sadi (kamayadi).
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Funksiyaning kritik va stasionar nuqtalari
1. y = f ( x) funksiyaning hosilasi nolga teng (ya`ni f ¢( x ) = 0 )
bo`lgan nuqtalar to`plamiga stasionar nuqtalar deyiladi.
2. y = f ( x) funksiyaning hosilasi mavjud bo`lmagan yoki nolga
teng (ya`ni f ¢( x ) = 0 ) bo`lgan nuqtalar to`plamiga kritik nuqtalar
deyiladi.
Funksiyaning maksimum va minimumlari
1. Funksiyaning maksimum va minimumlari nuqtalari shu
funksiyaning ekstremum nuqtalari, funksiyaning bu nuqtalardagi
qiymatlari esa funksiyaning ekstremumlari deyiladi.
2. Agar x0 nuqta y = f ( x) funksiyaning ekstremumi bo'lsa,
f ¢( x ) = 0 bo'ladi.
3. Funksiyaning maksimum va minimumlari:
x = x0 maksimum nuqtasi.
x = x0 minimum nuqtasi
Funksiyaning oraliqdagi eng katta va eng kichik
qiymati
1. y = f ( x) funksiyaning yopiq [ a, b ] oraliqdagi eng katta va eng
kichik qiymatlarini topish:
a) f ¢( x ) = 0 Þ xi Î [ a, b ] yoki xi Î [ a, b ] , i = 1, 2,3,... aniqlash;
b) agar xi Î [ a, b ] bo`lsa, f ( x1 ), f ( x 2 ), f ( x 2 ), ..., f ( a ), f ( b )
ni hisoblash;
v) agar xi Î [ a, b ] bo`lsa, f ( a ), f ( b ) ni hisoblash;
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g) bu qiymatlar ichidan eng kattasi va eng kichigi tanlab olinadi.
2. y = sin k x va y = cos k x funksiyalar uchun
max y = 1, min y = - 1.
3. y = a sin k x + b c o sk x funksiya uchun esa
max y =
a2 + b2 ,
min y = - a 2 + b 2 .
B O S H L A N G' I C H
FUNKSIYA
Agar berilgan oraliqdan olingan barcha x lar uchun F¢(x) = f ( x)
tenglik bajarilsa, u holda F ( x ) shu oraliqda f ( x ) funksiyaning
f ( x) Þ F ( x) + C deb
boshlang'ich funksiyasi deyiladi va
belgilanadi, C - ixtiyory o`zgarmas son.
Funksiyaning
1. C Þ Cx + C0 . 2. (kx ± b)n Þ
boshlang'ichlari
(kx ± b) n +1
1
+ C ( n ¹ -1) . 3. e kx ±b Þ e kx ±b + C.
k (n + 1)
k
1
1
Þ ln x + C.
5. sin (kx + b) Þ - cos (kx + b) + C.
x
k
1
1
6. cos (kx + b) Þ sin (kx + b) + C.
7. tg (kx + b) Þ - ln cos (kx + b) + C.
k
k
1
1
1
kx + b
8. ctg (kx + b) Þ ln sin(kx + b) + C.
9.
Þ ln tg
+ C.
k
sin(kx + b)
k
2
4.
10.
1
1
1
1
æ kx + b p ö
Þ ln tg ç
+ ÷ + C. 11.
Þ - ctg (kx + b) + C.
2
cos (kx + b)
k
2ø
sin (kx + b)
k
è 2
12.
1
1
1
Þ
+
+
tg
kx
b
C
(
)
.
13.
cos 2 ( kx + b )
k
x2 - a2
14.
1
1
x
Þ
arctg
+ C.
x 2 + a2
a
a
16.
18.
1
2
a + bx Þ
19. x + a
2
2
k x ±b
Þ ln x + x ± a + C. 17. a
2
x ±a
2
15.
2
3b
2
( a + bx )
3
Þ
1
a2 - x2
Þ arcsin
x
+ C.
a
ak x ±b
Þ
+ C, a > 0, a ¹ 1.
k × lna
+ C.
x
a2
2
2
Þ × x + a + ln x + x 2 + a 2 + C.
2
2
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+C.
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INTEGRALLAR
b
1. N'yuton-Leybnis formulasi:
S = ò f ( x ) dx = F ( x ) ba = F (b ) - F ( a ).
a
2. Egri chiziq bilan chegaralangan yuzalarni hisoblash:
b
a) Egrichiziqli trapesiya yuzi: S = ò f ( x ) dx ;
a
b) agar f1 ( x) > f 2 ( x) > 0 bo`lsa, u holda
b
S =
ò[f
1
( x ) - f 2 ( x ) ]d x ; bo`ladi.
a
3. y = f ( x) ( f ( x) > 0 ) egri chiziq
aylanganda hosil bo'lgan jism hajmi:
b
b
V = p ò f ( x ) dx = p ò y 2 dx.
2
a
a
AB=
: y f ( x), a £ x £ b yoyning uzunligi:
4. »
ì x = x(t ),
»
AB
:
í
5.
yoyning uzunligi:
î y = y (t ), a £ t £ b
b
l=ò
a
b
l=ò
a
1 + f ¢ 2 ( x ) dx .
x¢2 (t ) + y¢2 (t )dt .
6. y = f ( x) ( f ( x) ³ 0 ) , x Î [ a, b ] egri chiziqni OX o`qi atrofida
aylantirishdan hosil bo'lgan aylanish sirtining yuzini topish:
S
= 2p
b
ò
f (x) ×
1 + f ¢2 ( x ) d x .
0
Integrallash qoidasi
b
1.
ò
a
b
2.
b
k × f (x) dx = k
ò [ f (x) +
a
ò
f (x) dx,
k = co n st.
a
g ( x ) ]d x =
b
ò
b
f ( x )d x +
a
ò g ( x )d x.
a
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b
3.
ò
a
b
4.
ò
b
f ( x) d g ( x) = f (x) × g ( x)
b
a
-
ò g (x) d
f ( x ).
a
b
f ¢( kx + c ) dx = 1 k × f (kx + c ) a , k ¹ 0, c - o`zgarmas sonlar .
a
5. Agar f (-x) = f (x), x Î[ -a; a] , a > 0 bo`lsa,
a
ò
-a
a
f ( x ) dx = 2 ò f ( x ) dx
0
bo`ladi.
6. Agar f (-x) = - f (x), x Î[ -a; a] , a > 0 bo`lsa,
7. Agar f (x) ³ 0, x Î[ a; b] bo`lsa,
a
ò f (x)dx = 0 bo`ladi.
-a
b
ò
f ( x ) d x ³ 0 bo`ladi.
a
8. Agar a < x < c da f ( x ) ³ 0 ; c < x < b da f ( x ) < 0 bo`lsa,
b
ò
c
f ( x) dx =
a
ò
a
f ( x ) d x - ò f ( x ) d x bo`ladi.
a
c
Aniqmas integral
1.
3. ò
dx
ò x × lnx = ln ln x + C.
x dx
= - 1 - x + C. 4.
2
1 - x2
2.
a
òx
m
ò sin x × cosxdx =
a +1
× ln xdx = x
1
sin m +1 x + C.
n +1
æ ln x
1 ö
×ç
÷ + C (a ¹ -1) .
ç a +1 (a +1)2 ÷
è
ø
x 2
a2
x
2
5. ò a - x dx =
a - x + arcsin + C .
2
2
a
1
6. ò arctgx dx = x × arctgx - × ln (1 + x 2 ) + C .
2
7. ò x × e x dx = ( x - 1) × e x + C . 8. ò x 2 e x dx = ( x 2 - 2 x + 2 ) × e x + C .
2
2
x 1
- sin 2 x + C .
2 4
cos3 x
3
11. ò sin x dx = -cos x +
+ C.
3
9.
13.
2
ò sin xdx =
ò ln
a
x 1
+ sin 2 x + C .
2 4
sin3 x
3
12. ò cos x dx = sinx + C.
3
10.
2
ò cos xdx =
xdx = x × ln a x - a ò ln a -1 x dx + C .
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14.
ò arcsin x dx =x × arcsin x +
1 - x + C.
2
2
2cx + b
ì
2
,
+
< 4ac,
arctg
C
agar
b
ï
2
2
4ac - b
dx
ï 4ac - b
15. ò
=
í
a + bx + cx2 ï
1
2cx + b - b2 - 4ac
+ C, agar b2 > 4ac.
ln
2
2
ï b - 4ac 2cx + b + b - 4ac
î
dx
1
ln 2cx + b + 2 c a + bx + cx 2 + C , c > 0.
16. ò
=
c
a + bx + cx 2
17.
ò
a + bx + cx 2 dx =
-
1 8.
19.
20.
21.
22.
23.
24.
25.
26.
ò
b 2 - 4 ac
8 c2
2cx + b
a + bx + cx 2 4c
ln 2 cx + b + 2 c
dx
=
a + bx - cx 2
1
b + 4ac
2
ln
a + b x + cx 2 + C .
2 cx - b +
- 2 cx + b +
b 2 + 4 ac
b + 4ac
2
+ C.
1
2cx - b
arcsin
+ C , c > 0.
ò a + bx - cx 2 c
2
b + 4ac
2cx - b
2
2
a
+
bx
cx
dx
=
a
+
bx
cx
+
ò
4c
2cx - b
b2 + 4ac
+
arcsin
+ C.
2
2
8 c
b + 4ac
a+x
ò b + x dx = ( a + x )( b + x ) + ( a - b ) ln a + x + b + x + C.
a-x
a+x
dx
=
a
+
x
b
+
x
+
a
+
b
arcsin
+ C.
(
)(
)
(
)
ò b+ x
a+b
a+x
b-x
dx
=
a
+
x
b
x
a
+
b
arcsin
+ C.
(
)(
)
(
)
ò b-x
a+b
ò sh xd x = ch x + C ,
ò ch xd x = sh x + C .
dx
=
(
ò thxdx =
lnchx + C ,
ò sin mx × sin nx dx = -
ò cthxdx =
)
lnshx + C .
sin ( m + n ) x sin ( m - n ) x
+
+ C , m ¹ n.
2 (m + n)
2 (m - n)
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27.
ò cos mx × cos nx dx =
28.
ò sin mx × cos nx dx = -
29.
òe
òe
30.
sin ( m + n ) x sin ( m - n ) x
+
+ C , m ¹ n.
2 (m + n)
2 (m - n)
cos ( m + n ) x cos ( m - n ) x
+ C , m ¹ n.
2(m + n)
2 (m - n)
ax
× sin nx dx = e ax ( a × sin nx - n × cos nx ) ( a 2 + n 2 ) + C .
ax
× cos nx dx = e ax ( a × sin nx + n × cos nx ) ( a 2 + n 2 ) + C .
ì
ï
ï
dx
ï
= í
31. ò
a + bcosx
ï
ï
ï
î
ì
ï
ï
ïï
dx
=
32. ò
í
a + b s inx
ï
ï
ï
ïî
æ a -b
xö
× tg ÷÷ + C, agar a > b,
arctg çç
2ø
a 2 - b2
è a+b
x
b - a × tg + a + b
1
2
ln
+ C , agar a < b.
2
2
x
b -a
b - a × tg - a + b
2
x
a × tg + b
2
2
arctg
+ C , agar a > b,
2
2
2
2
a -b
a -b
x
a × tg + b - b2 - a 2
1
2
ln
+ C , agar a < b.
2
2
x
2
2
b -a
a × tg + b + b - a
2
2
KOMBINATORIKA
ELЕMЕNTLARI
1. m ta elеmеntdan n tadan barcha o`rinlashtirishlar soni:
m!
Amn = m(m - 1)(m - 2)...(m - n + 1) =
, bu еrda m! = 1× 2 × 3 × ... × m .
(m - n)!
2. n ta elеmеntdan barcha o`rin almashtirishlar soni:
Pn = n ! = 1 × 2 × 3 × ... × n .
3. m ta elеmеntdan n tadan barcha gruppalashlar soni:
Сmn
=
Amn
Pn
=
m!
,
n!( m - n)!
Cm0 = Cmm = 1 .
4. N`yuton binomi:
n
( a + x ) = C n0 a n + Cn1 a n -1 x + C n2 a n - 2 x 2 + ... + Cnk a n - k x k + ... + Cnn x n
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Q Y S H I M CH A
M A` L U M O T L A R
1. Ketma-ket kelgan sonlar ko`paytmasi (1× 2 × 3 × ××× × n = n !) nechta
énù é n ù é n ù
x
=
nol bilan tugashi:
êë 5 úû + êë 52 úû + êë 53 úû + ... .
2. a - soat va minut strelkalari orasidagi burchak, b - t vaqtdan
2
t
=
3600 - (a + b ) ) bo`ladi.
(
keyin ular orasidagi burchak bo`lsa,
11
3. lg 2 » 0,3010, lg 3 » 0, 4771,
lg 300 + lg 8 = lg 3 + lg10 2 + lg 23 » 0, 4771 + 2 + 3 × 0,3010 = 3,3801.
4. Funksiyaning Teylor formulasi: Agar y = f ( x ) funksiya [ a, b]
( j)
kesmada berilgan bo`lib, x0 Î( a, b) nuqtada f (x0 ), ( j =1, 2, ..., n +1)
hosilalar mavjud bo`lsa, u holda
f ¢( x 0 )
f ¢¢( x0 )
( x - x0 ) +
( x - x0 ) 2 + ... +
1!
2!
(n )
( x0 )
f
+
( x - x 0 ) n + R n ( x ),
n!
f ( x ) = f ( x0 ) +
f ( n +1) ( x0 + q ( x - x0 ) )
( x - x0 )n +1 , 0 < q < 1 bu erda Rn ( x ) =
( n + 1)!
Teylor formulasining qoldiq hadi.
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