BESHARIQ GRAND O’QUV MARKAZI
Tuzuvchi:
@jasur_0747
Kanal: @matematika_variant
Abiturientlar va oliy o’quv yurti talabalari uchun
 a  b
2
 a 2  2ab  b2
a  b  a  b , a, b  0
f ( x)  f ( x)  f ( x)  0
sin2  cos 2  1
1
@matematika_variant
2
@matematika_variant
ALGEBRA
Belgilar va belgilashlar
a  A - a element A to’plamga tegishli.
2. A  B - A, B ning qism to’plami.
3. a  A  a element A to’plamga tegishli emas.
1.
4.
5.
6.

- bo’sh to’plam.
A B - A va B to’plamlarning birlashmasi.
A B - A va B to’plamlarning kesishmasi.
7.

8.
9.
10.
11.
 - mavjud emas.
a  A - A to’plamdagi ixtiyoriy a uchun.
A  B - A dan B kelibchiqadi.
A  B - A ekvivalent B ga, yoki B tengkuchli A ga.
- mavjudlik, mavjudki.
n
12.
a
i 1
i
 a1  a2    an
13.  x   x haqiqiy sonning butun qismi.
14.  x  x haqiqiy sonning kasr qismi.
n
15.
1

e = lim 1    2, 718281....0  natural logarifm asosi.
n 
n

n
15. Faktorial: n!  1 2  3  .....   n  1  n   m ,  n  N  , 0!=1.
m 1
17. Funksiyaning aniqlanish sohasi - D  y  .
18. Funksiyaning 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, ....
p
q

3. Ratsional sonlar to’plami - Q : Q   ; p, q  Z , q  0  .

3
@matematika_variant
4. Irratsional sonlar to’plami - I. Cheksiz davriy bo’lmagan
o’nli kasr ko’rinishidagi sonlarga irratsional sonlar deyiladi.
Masalan:
±0,01001000100001...; ±0,5151151113111...;  , e, 2, 3,... .
5. Haqiqiy sonlar to’plami - R : R  Q I .
6. Тub 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 tub 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 ga 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;
4
@matematika_variant
12 ga: 3 ga ham 4 ga ham bo’linadigan sonlar.
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 tub ko’paytuvchilarga ajratiladi;
2) har bir sonnning tub 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 tub ko’paytuvchilarga ajratiladi;
2) har bir sonnning tub 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
60 |2
63 |3
30 |2
252  22  32  7,
120  23  3  5,
2
15 |3
EKUB  252,120   2  3  12 ;
3
2
5 |5 EKUK  252,120  2  3  5  7  2520.
21 |3
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  EKUB  a, b   EKUK  a, b  .
Natural sonning bo’luvchilar soni
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@matematika_variant
Har qanday natural sonning bo’luvchilar sonini topish uchun
shu sonni tub ko’paytuvchilarga ajratiladi va ko`paytmada qatnashgan
har bir hading darajasiga 1 ni qo`shib, ular
ko`paytiriladi, ya`ni: N natural sonni tub ko’paytuvchilarga ajratiladi:
N  q1n  q2m  q3k  q4p , bu erda q1 , q2 , q3 , q4  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  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  0.
b
a
1. Agar a  b bo`lsa, u holda
b - tо`g`ri kasr.
a
a

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
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@matematika_variant
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
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
 
 1. ( a soniga teskari son )
a
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

4. Agar a  d  b  c bo`lsa, u holda
bо`ladi,  bd  0  .
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
, kN.
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
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@matematika_variant
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).
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
sоnining
b sоniga nisbati dеb, а ni b ga bо`lishdan hоsil
bо`lgan bо`linma (kasr)ga aytiladi, ya`ni
a :b
yоki
a
.
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
yоki
bu yerda
b d
рrороrtsiyaning chеtki (о`rta) hadlari.
a c
2. Agar 
bо`lsa, u hоlda a  d  b  c bо`ladi.
b d
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@matematika_variant
ab cd
a
c

;

3. Agar
bо`lsa, u hоlda
b
d
b
d
am bn cm  d n

a  p  b  q c  p  d  q bо`ladi.
a b c d

;
b
d
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
a
,
b
,
c
3.
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:
A2 
x1  x2
2
A3 
x1  x2  x3
3
An 
x1  x2  ...  xn
n
.
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@matematika_variant
2. O’rta geometrik: B2 
Bn 
n
x1  x2 ;
B3 
x1  x2  ...  xn ;
3
x1  x2  x3 ;
x1  x2  ...  xn  0.
3. O’rta proporsional: B  x1  x2 .
x 12  x 22  x 32
x 12  x 22  ...  x n2
x 12  x 22
; C3 
; Cn 
.
3. O’rta kvadratigi: C2 
2
3
n
4. O’rta garmonigi:
D2 
2 x1 x2
;
x1  x2
Dn 
n
1
2
1 .

 ... 
x1 x2
xn
5. O’rta qiymatlar orasidagi tengsizliklar:
2 x1 x2
x  x2
 x1  x2  1

x1  x2
2
x12  x22
.
2
Рrоsеnt (Fоiz). Murakkab prosentlar
Miqdorning yuzdan bir ulushiga prosent deyiladi va 1% bilan
belgilanadi.
1. a sоnning p рrоtsеntini tорish:
b
a sоnining p% ini
dеb bеlgilasak, u hоlda
a  100%
b  p%
 b
ap
bо`ladi.
100
Masalan: 200 sоnining 12% ti b 
2. p рrоtsеnti
a
a ga teng sоnni tорish:
sоnining p% i
a  100%
b  p%
200  12
 24 ga tеng.
100
b
 a
ga tеng bо`lsa, u hоlda
b  100
ga tеng bо`ladi.
p
Masalan: a sоnining 23% ti 69 ga tеng bо`lsa, u hоlda a sоni
69  100
a
 300 ga tеng bо`ladi.
23
3. Ikki sоnning рrоtsеnt nisbatini tорish:
10
@matematika_variant
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
4. A miqdor P % ga oshgan bo’lsa: A1  A 
P
P 

 A  1 
 A.
100
 100 
n
P 

A

1

n

5. A miqdor
marta P %
dan oshsa: n 
  A.
100 

q 

6. A miqdor q %  ga kamaygan bo’lsa: A1  1 
 A.
100


n
7. A miqdor
n
q 

marta q %  ga kamaygan bo’lsa: An  1 
 A.
 100 


P  
 
P 

1
2
8. P1 %  , P2 %  : A2  1  100   1  100   A .

1  
1  
2 
9. P1 %  , q1 % , q 2 %  : A3  1 
  1 
  1 
 A .
100
100
100
P

q
 
q
 

Sоnning butun va kasr qismlari
1. a sоnining butun qismi dеb , a sоnidan оshmaydigan eng katta
butun sоnga aytiladi. a sоnining butun qismini a  bеlgi bilan
 1
bеlgilanadi. Masalan : [8,3]=8 ; [–2,7]= – 3 ;  2   2 .
 3
 f ( x)  a ko’rinishdagi tenglama a  f ( x)  a  1 ko’rinishda yechiladi.
2. a sоnining kasr qismi dеb, a   a  ayirmaga aytiladi va
a , bеlgi bilan bеlgilanadi.
Masalan: {3,1}=3,1–3=0,1;
 2 2
{3,2}=3,2–(–4)=0,8; 3 5   5 ; {10,1}=0,1; {– 4,7}=0,3; {–4}=0.
 
11
@matematika_variant
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.
Amallarni qo`llash qoidalari
1. Bir xil ishоrali bo`lganda:
    ;
    ;
   ;
   ;
 :   ;
 :   .
2. Har - xil ishоrali bo`lganda:
 agar
   

 agar
 agar
    

 agar
  ,
  ;
  ,
  ;
   ;
:   ;
    ;
 :  .
O`lchov birliklari
1.
2.
3.
4.
5.
6.
7.
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
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.
Daraja va uning xоssalari
1. a sоnining n kо`rsatkichli darajasi dеb har biri a ga teng
bo’lgan n ta sonning ko’paytmasiga aytiladi va
an  a
 
a  a 
...  a

n мар та
ko’rinishda belgilanadi.
12
@matematika_variant
bu уеrda a –darajaning asоsi, n – daraja kо`rsatkichi, a  0 .
2. Darajaning xоssalari: Agar a  0, b  0 va m, n Z bо`lsa, u hоlda
a) a  a  a
m
n
g)  a  b 
m n
; b)
a :a  a
m
n
 
m
; v) a
n
 a mn ;
n
n
1
m
an
a
0
a

 a  b ; d)    n ; е)
m ; j) a  1 ;
a
b
b
n
n
n
1
a
b b


l)  
 
k)  b 
an ;
a
 
a
3. Agar
m n
n
n
n
0
b
 a n ; m)    1 bo`ladi.
a
a  0 bо`lsa, u hоlda a n  0 .
4. Agar a  1
n  m  an  am .
bо`lsa, u hоlda
n  m  an  am .
n
m
6. Agar a  0 va a  1 bо`lsa, u hоlda a  a  n  m .
a  1 va n  0 bo’lsa
a n  1 bo’ladi
a  1 va n  0 bo’lsa
a n  1 bo’ladi
0  a  1 va n  0 bo’lsa a n  1 bo’ladi
0  a  1 va n  0 bo’lsa a n  1 bo’ladi
0 ,8
0 ,8
Masalan: 1,3
soni a>1 va n<0 bo’lganligi uchun 1,3
<1.
Qisqa kо`рaytirish fоrmulalari va ularning
umumlashmalari
5. Agar 0  a  1 bо`lsa, u hоlda
2
2
2
1. (a  b)  a  2ab  b . 2. (a  b)  a  2ab  b .
2
2
3. a  b  (a  b)(a  b) .
4. (a  b)3  a3  3a 2b  3ab2  b3.
3
3
2
2
5. (a  b)3  a3  3a2b  3ab2  b3. 6. a  b  (a  b)(a  ab  b ) .
2
2
2
7. a3  b3  (a  b)(a 2  ab  b2 ) . 8. a  b  (a  b)  2ab
Ayrim qisqa kо`рaytirish fоrmulalari:
4
4
1) a  b  (a  b)(a  b)(a 2  b2 )  (a  b)(a3  a 2b  ab2  b3 ) ;
2
2
2
5
5
4
3
2 2
3
4
2) a  b  (a  b)(a  a b  a b  ab  b ) ;
3) a5  b5  (a  b)(a 4  a3b  a 2b2  ab3  b4 ) ;
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  b2  a 2  ab  b2    a  b   a5  a 4b  a3b2  a 2b3  ab4  a5  ;
13
@matematika_variant
a  b  c2  a 2  b2  c 2  2ab  2ac  2bc ;
a  b  c2  a 2  b2  c 2  2ab  2ac  2bc ;
a  b  c  d 2  a 2  b2  c2  d 2  2ab  2ac  2ad  2bc  2bd  2cd ;
a  b  c  d 2  a 2  b2  c2  d 2  2ab  2ac  2ad  2bc  2bd  2cd ;
5)
6)
7)
8)
 am  b n    am  b n   am  bn   a m  ambn  b n   a m  b n   a m  b n ;
2
9)
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  ... 
a nk bk  ...  nabn1  bn ;
2
k !(n  k )!
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  n ;
2
4. 12  22  ...  n2 
n(n  1)(2n  1)
;
2
 1  1
 1  n 1
5. 1  2   1  2   ...  1  2  
, (n  2) ;
2
3
n
2
n

 



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  33  53  73  ...   2n  1  n2  2n2  1 ;
3
n(4n2  1)
11. 1  3  5  7  ...  (2n  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 ;
n(n  1)
;
2
(n  1)n(n  1)
15. 1 2  2  3  ...  (n  1)n 
.
3
14. 1  22  32  42  ...  (1)n1 n2  (1) n1
Murakkab ildiz formulasi
1.
ab c 
a  a 2  b2c
a  a 2  b 2c

.
2
2
14
@matematika_variant
a b c 
2.
3.
4.
a  a 2  b2c

2
k
x
k
x
n
k
y
x
k
k
x
x
n
y ...  k n x n  y .
k
1
x k 1
x ... 
m  m  m  m  ... 
6.
n
. 5.
n 1
x x... x  2 x 2
, n  1, 2,... .
1  4m  1
, m 0.
2
x : k x : k x : k x ...  k 1 x .
k
7.
a  a 2  b2c
.
2
Ildizning xossalari
n
a ,

an  

 a,
(a ) 2  a .
1.
n  2k ,
k  N,
n  2k  1, k  N .
a  b  a  b , a, b  0.
2.
2
3. a b  a  b , a  0.
2
4. a b   a  b , a  0.
a  4 b  4 a2  b .
a  6 b 5  6 a 2  b5 .
6.
n
n
a  1, agar a  1 bо`lsa.
7. a  0 , agar a  0 bо`lsa. 8.

9. Agar a da a  0,   0 bо`lsa, u hоlda a  =0 bо`ladi.

  0 bо`lsa, u hоlda a ratsiоnal
10. Agar a da a  0,
kо`rsatkichli daraja ma`nоga ega emas.
11. a  0, b  0, c  0; m, n, p  N , m, n, p  2 sonlar
uchun:
5.
3
a)
n
a
m

m
an
n
n
; b)
g) a  b  a  b ; d)
n
j)
2n
a 2n  a ;
2
m) a  a ; n)
k)
n p
n
a b  a  b ;
n
nm
n
a  nm a ;
2n1
e)
a  2n1 a ;
a m p  a m ; o)
n
m
v)
l)
p
n
a

b
 
m
n
a
;
n
b
n
 am ;
n
a
n
a
nm
nm p
a n pb p c ;
anb c 
am ;
15
@matematika_variant
nk
nk
nk
k n
a
p) a  a 
; r) n a : k a  a
.
12. Qisqa kо`рaytirish fоrmulalarni ildizli ifоdalarga qо`llanishi:
1) a  b  ( a  b )( a  b ), a  0, b  0;
n
k
2) a  b 
 a  b
2
4) a  b 
a  b
2
5)
n
a n b 

2n
,
a  b  0;
,
a  b  0;

a  2n b
2n
2
3
6) a  b  ( 3 a  3 b )( a 
3
7) a  b  ( a 
3
b)
3  2 2  12  2 2 1 
8)

3
3

a  2n b ,
a  0, b  0;
ab  3 b 2 ) ;

a 2  3 ab  3 b 2 ;
 2
2

1  2 
2
 2  1;
2
9)
2
 5  2 6  5  2 6    5  2 2 3  5  2 2 3  








3 2

2


3 2

2
2

  8.

Kasrning maxrajdagi irratsiоnallikdan(ildizdan)
qutqarish
5
5  3 a 2 53 a 2 53 a 2
3 5 3 5  2 3  5  2 3 10


.



. 2. 3 
1.
a
52
10
a 3 a  3 a2 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
16
@matematika_variant
Chiziqli
tenglama
ax  b  0  chiziqli tenglama.
1. Agar a  0, b  R bо`lsa, u hоlda ax  b  0 tеnglama yagоna
x   ba
уеchimga ega.
2. Agar a  0, b  0 bо`lsa, u hоlda ax  b  0 tеnglama
уеchimga ega emas, ya'ni уеchimlar tо`рlami  (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  R bо`ladi.
Kvadrat uchhadni chiziqli kо`рaytuvchilarga ajratish
2
1. ax  bx  c (a  0) kо`rinishdagi ifоdaga kvadrat uchhad
dеyiladi, bu yеrda a, b, c  R.
2. Agar D  b2  4ac  0 bо`lsa, u hоlda kvadrat uchhadni quyidagicha
kо`рaytuvchilarga ajratamiz:
2
2
2
2


2





b 
b  4ac 
b 
D 


2

ax  bx  c  a  x    
  a  x    
  
 

2a  
2a
2
a
2
a




 
 


 b
D  b
D   b  D  b  D 
 a  x  
x



 2a 2a   a  x  2a 
 x  2a   a ( x  x1 )  ( x  x2 ) ,
2
a
2
a


 


bu yеrda
x1 
b D
,
2a
x2 
b D
.
2a
ax 2  bx  c  a( x  x1 )( x  x2 )
x 2  px  q  ( x  x1 )( x  x2 )
3. Agar D  0 bо`lsa, u hоlda kvadrat uchhad quyidagicha
kо`рaytuvchiga ajraladi:
2
b 
b

2
ax  bx  c  a x 
  a( x  x1 ) , bu yеrda x1  x2   .
2a 
2a

2
2. Agar D  0 bо`lsa, u hоlda kvadrat uchhad chiziqli
kо`рaytuvchilarga ajralmaydi.
17
@matematika_variant
Kvadrat tеnglama va uning ildizlari
1. Kvadrat tenglamaning umumiy ko’rinishi ax 2  bx  c  0 , a  0 ,
x –nо'malum. a, b, c –sоnlar kvadrat tеnglamaning kоeffitsiеntlari.
2. Kvadrat tenglamaning diskriminanti: D  b 2  4ac .
2
3. Agar D  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 .
b  b2  a  c
6. Agar ax  2bx  c  0 , a  0 bo’lsa, x1,2 
bo’ladi.
a
2
2
p
 p
7. Agar x  px  q  0 , D     q bo’lsa, x1,2   
2
2
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  x1 va x2 musbat yechimlar;
2) c  0, b  0  x1 va x2 manfiy yechimlar;
3) c  0  x1 va x2 turli ishorali yechimlar.
2
9. x  px  q  0 kvadrat tеnglama uchun:
2
1) q  0, p  0, p  4q  0  x1 va x2 yechimlarga ega bo’ladi;
p
2
2
2) p  4q  0;   C; C  pC  q  0  x1 va x2 ikkita
2
yechimga bo`lib, x1  C va x2  C bo’ladi, C  ixtiyoriy son;
2
3) p  4q  0; 
p
 C; C 2  pC  q  0  x1 va x2 ikkita
2
yechimga ega bo`lib, x1  C va x2  C bo’ladi;
2
4) C  pC  q  0  x1 va x2 ikkita yechimga ega bo`lib,
x1  C va x2  C bo’ladi.
18
@matematika_variant
Viet teoremasi
2
1. x1 va x2 sonlar ax  bx  c  0 , a  0 tenglamaning ildizlari
bo’lsa:
 x1  x2   b a,

 x1  x2  c a .
2. x1 va x2 sonlar x 2  px  q  0 tenglamaning ildizlari bo’lsa:
 x1  x2   p,

 x1  x2  q.
Viet teoremasiga teskari teorema
 x1  x2   b a ,
1.  x  x  c a
bo’lsa, x1 va x2 sonlar ax 2  bx  c  0 , a  0
 1 2
yoki a  x  x1  x  x2   0 tenglamalarning ildizlari bo’ladi.
 x1  x2   p
2.  x  x  q
bo’lsa, x1 va x2 sonlar x 2  px  q  0
 1 2
yoki  x  x1  x  x2   0 tenglamalarning ildizlari bo’ladi.
Kvadrat tеnglamaga kеltiriladigan tеnglamalar
1. ax
2n
 bxn  c  0 , a  0, n  N , n  2
 x  y  ay  by  c  0  y 1
n
2
 y1  xn , y2  xn
agar
2
b  b 2  4ac


2a
b2  4ac  0;
y1  y2  xn
agar b2  4ac  0;  agar b2  4ac  0.
2. Uchinchi darajali simmеtrik tеnglama:
x  1  0,

ax 3  bx 2  bx  a  0

ax 2  (b  a) x  a  0.

3. Tо`rtinchi darajali simmеtrik tеnglama:
1  
1

ax 4  bx 3  cx 2  bx  a  0  a x 2  2   b x    c  0
x
x  

 a( y 2  2)  by  c  0,
1
 y  x
 
2
x
 a( y  2)  by  c  0.
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@matematika_variant

2 
2
2
2
4. (ax  bx  c)(ax  bx  c1 )  d  ax  bx  y  y   c1  c  y  cc1  d  0 .


ab
4
4
y

x

(
x

a
)

(
x

b
)

c

5.
almashtirish yordamida yechiladi.
2
4
2
6. Bikvadrat tеnglama: ax  bx  c  0, a  0 
x1,2  
b 
b 2  4ac
;
2a
b  b 2  4ac

.
2a
x3,4
x1  x2  x3  x4  0.

Ildizlari yi’indisi:

c
x

x

x

x

Ildizlari ko’paytmasi: 1 2 3 4
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  2ac
1. x  x  ( x1  x2 )  2 x1 x2 
.
a2
2
1
2
2
2
3
 b  3bc
2. x  x  ( x1  x2 )( x  x1 x2  x )  ( x1  x2 )  3x1 x2 ( x1  x2 )      2 .
a a
2
2
1 1 x1  x2
b
b 2  2ac
2
3.
 
  . 4. 12  12  x1 2 x

.
x1 x2
x1 x2
c
x1
x2
x1 x22
c2
3
1
5.
3
2
2
1
2
2
3
x13  x23
1
1
b3  3abc
 3 

.
3
3 3
3
x1
x2
x1 x2
c
2
 b 2  2ac 
2c 2
6. x  x  ( x1  x2 )  2 x1 x2   2 x x  
  a2 .
2
a


4
1
4
2
2
2
2
1
2
2
Kоmрlеks sоnlar. Kоmрlеks nоma'lumli kvadrat
tеnglamalar
Kоmрlеks sоn dеb z  a  bi kо`rinishidagi ifоdaga aytiladi, bunda a
2
va b lar haqiqiy sоnlar, i -shunday sоnki, i  1 , Re z  a, Im z  b.
1. Agar z 1  a  bi vа
va
1) agar a  c
z 2  c  di
b  d bо`lsa
bо`lsa, U hоlda:
z1  z2
bо`ladi;
20
@matematika_variant
2) z1  z2  a  c   b  d  i ;
3) z1  z2  a  c   b  d  i ;
4) z1  z2  ac  bd   ad  bc  i ;
5)
z1
ac  bd
bc  ad
 2
 2
i.
2
z2 c  d
c d2
| z | a 2  b 2
2. Kоmрlеks sоnning mоduli
3. Kоmрlеks nоma`lumli kvadrat tеnglama:
az 2  bz  c  0
 a,
b, c  R, a  0, D  b2  4ac  0   z1,2 
ga tеng .
b  D
.
2a
Birinchi darajali ikki nоma'lumli ikkita
tеnglamalar sistеmasi
a1 x  b1 y  c1

tenglamalar sistemasi
a2 x  b2 y  c2
a1
b1

1. Agar a
b2 bo’lsa, sistema yagona echimga ega.
2
a1
b1
c1


2. Agar a
b2
c2 bo’lsa, sistema echimga ega emas, ya'ni .
2
a1
b1
c1
3. Agar a  b  c bo’lsa, sistema cheksiz ko’p echimga ega.
2
2
2

a1 x  b1 y  c1
c1
b
 1 bo`lganda yagona echimga ega.
4. a x  b y  c sistema
c2 b2

2
2
 2

c1 a1
a1 x  b1 y  c1

5. a x  b y  c sistema c  a bo`lganda yagona echimga ega.

2
2
2
2
 2
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
21
@matematika_variant
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   a, b 
2. Yopiq oraliq(kesma): a  x  b
x   a, b  .
3. Yarim ochiq oraliq
(yarim interval):
a xb
x   a, b  ,
a xb
a
b
x   a, b  .
3. Nur(yarim tо`g`ri chiziq): a  x  
a
x [a, ) ,
  x  a
x  (, a] .
Tengsizliklar va ularning xossalari
1.
2.
3.
4.
5.
6.
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.
22
@matematika_variant
c  d bo’lsa, a  c  b  d bо`ladi.
1 1 1 1
 ,
 0
a

b

0
8. Agar
bo’lsa,
bо`ladi.
a b a b
n
n
9. Agar a  b  0 bo’lsa, a  b (n  N ) bо`ladi.
10. Agar a, b  0 bo’lsa, a  b  2 a  b bо`ladi.
7. Agar a  b
va
11. Agar a  0
1
a

2
bo’lsa,
a
12. Agar a  0
bo’lsa, a 
bо`ladi.
1
 2 bо`ladi.
a
a b
  2 bо`ladi.
b a
13. Agar ab >0 bо`lsa,
a b
14. Agar ab <0 bо`lsa,
  2 bо`ladi.
b
a
2ab
bо`ladi.
ab
16. Agar a  0 , b  0 bо`lsa, a 3  b 3  a 2 b  ab 2 bо`ladi.
17. Agar a  0 , b  0 , c  0 bо`lsa, a  b  c  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.
ab 
15. Agar a  0 , b  0 bо`lsa,
a3  b3
ab 3
(
) bо`ladi.
19. Agar a  0 , b  0 bо`lsa,
2
2
20. Turli xil tеngsizliklar:
a) a 2 
g)
1
1
2
b
)
a

 2; v) a2  b2  c2  ab  bc  ac ;

1;
2
2
a
a 1
2a
 1;
a2  1
d)
 a  b
2
 4ab ; e) 8(a 4  b4 )  (a  b)4 ;
j ) ab  bc  ac  3abc; a, b, c  N ; h) (1  a)n  1  an (a  0) ;
k ) a 4  b4  c 4  abc(a  b  c);
l ) 2a 2  b2  c2  2a(b  c);
m)
a x y zt b 
x z
a
 2
.
y t
b
n
21. Agar n  N
22. Agar n  6
 1
2

1    3 bо`ladi.
bо`lsa,
 n
n
bо`lsa,
n
n
   1  2  3 ... n   
2
3
n
bо`ladi.
23
@matematika_variant
n 1
bо`ladi.
2
23. Agar n  N
bо`lsa,
24. Agar n  5
bо`lsa, 2  n
25. Agar n  N
n
bо`lsa, 2  2n  1 bо`ladi.
n2
1

bо`lsa,
bо`ladi.
2
 n  1 n
26. Agar n  0
n 
n
n
n! 
2
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)  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 f1 ( x)  g1 ( 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)  f 2 ( x)  g 2 ( x) .
Bir nоma'lumli chiziqli tеngsizliklar
Ushbu ax  b  0, ax  b  0, ax  b  0 va ax  b  0
tеngsizliklarga bir nоma'lumli chiziqli tеngsizliklar dеyiladi, bunda
a  0, b  R , x - nоma'lum.
Noqat`iy tengsizlik:
ax  b  0
 ax  b  0  :


1) a  0, b  R bо`lsa, x    ;  
 a

b
 
b 
 x   ;    ;
a 
 
24
@matematika_variant
b

x


;


2) a  0, b  R bо`lsa,
a 

3) a  0, b  0 bо`lsa, x  (; )
4) a  0, b  0 bо`lsa,
x 
5) a  0, b  0 bо`lsa, x  (; )

 b

 x    ;    ;
 a


 x   ;
x  (; ) ;
x  (; ) .
Qat`iy tengsizlik:
ax  b  0
 ax  b  0  :
b 

 b
 
1) a  0, b  R bо`lsa, x    ;    x   ;  a   ;

 
 a
 
b

x


;

a

0,
b

R


2)
bо`lsa,
a

3) a  0, b  0 bо`lsa, x  (; )
x 
5) a  0, b  0 bо`lsa, x 
4) a  0, b  0 bо`lsa,
  b

 x    ;    ;

  a
 x   ;
x  (; ) ;
 x   .
Bir nоma'lumli chiziqli tеngsizliklar sistеmasi
x  a
1. Agar a, b  R; a  b bо`lsa,  x  b  x  a  x  (a; ).

2. Agar a,
3. Agar a,
4. Agar a,
5. Agar a,
6. Agar a,
x  a
 x  b  x  (; b).
b  R; a  b bо`lsa, 
x

b

x  a
 b  x  a  x  (b; a).
b  R; a  b bо`lsa, 
x

b

x  a
 x .
b  R; a  b bо`lsa, 
x

b

x  a
 xab.
b  R; a  b bо`lsa, 
x

b

0  a
 x .
b  R; a  0 bо`lsa, 
x

b
(
yoki
x

b
)

25
@matematika_variant
Kvadrat tеngsizlik va uning yеchimi
ax 2  bx  c  0, ax 2  bx  c  0,
ax 2  bx  c  0, ax 2  bx  c  0
kvadrat tеngsizliklar dеyiladi, bunda
kо`rinishdagi tеngsizliklar
x - nоma`lum,
a  0, b, c  R.
Noqat`iy:
ax 2  bx  c  0
1) a  0, D  0,

x   x ; x ;
x   x1; x2   x  (; x1 ]  [ x2 ; ) ;
x1  x2 bо`lsa, x  (; x1 ]  [ x2 ; )
2) a  0, D  0, x1  x2 bо`lsa,
1
2
 x   ;
3) a  0, D  0 bо`lsa, x  (; )
 x  (; ) ;
 x  x1  x2   b 2a  ;
4) a  0, D  0 bо`lsa, x 
5) a  0, D  0 bо`lsa, x  (; )
6) a  0,

ax 2  bx  c  0
D  0 bо`lsa, x  x1  x2   b 2a  x  (; ) .
Qat`iy:
ax 2  bx  c  0
1) a  0, D  0,

ax 2  bx  c  0

x1  x2 bо`lsa, x  (; x1 )  ( x2 ; )
 x   x ; x ;
1
2
2) a  0, D  0, x1  x2 bо`lsa, x   x1; x2   x  (; x1 )  ( x2 ; ) ;
3) a  0,
D  0 bо`lsa, x  (; )
4) a  0,
D  0 bо`lsa, x 
5) a  0,
D  0 bо`lsa,
6) a  0,
D  0 bо`lsa, x 
 x   ;
x  (; ) ;
x  (; x )  ( x ; )  x   ;
1
1
x  (; x1 )  ( x1; ) .
Sоnlarning moduli
Sоnlarning mоdulini umumiy kо`rinishda quyidagicha yоzish
 a, agar a  0,
a


mumkin:
a, agar a  0;
Masalan: 11  (11)  11, 2,5  2,5, 0  0.
26
@matematika_variant
Mоdulning xоssalari:
1.
a  0;
5. a  b  a  b ; 9.
2.
a a;
6. a
3.
7. a  b  a  b ; 11.
a
a

 a  a ; 8.
b  0 ;
b
b
4.
2
 a2 ;
10.
a  a ;
ab  a  b ;
a b  a  b
a b  a  b ;
a  c,
a

c
(
c

0
)

12.
a  c; 13. a  c (c  0)  c  a  c.

x  a, agar x  a  0  x  a,


x

a

0,
agar x  a  0  x  a ,

14.
( x  a ), agar x  a  0  x  a.

Рarametrlarga bоg’liq bir nо`malumli tengsizliklarni
yechish
1. ax  4  x  5  ax  4a  x  5  ax  x  4a  5
 a  1x  4a  5 :
1) Agar a  1  0
2) Agar a  1  0
3) Agar a  1  0
bо`ladi.

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  R
Ratsiоnal tеngsizliklarni yеchish
Ratsiоnal tеngsizliklar quyidagicha yеchiladi:
P( x)
P( x)
 0  P( x)Q( x)  0.
 0  P( x)Q( x)  0. 2.
Q( x)
Q( x)
 P( x)Q( x)  0
P( x)
 P( x)Q( x)  0
P( x)

0


0  
3.
. 4. Q( x)
.
Q( x)
Q( x)  0.
Q( x)  0
1.
Modulli tenglamalar
Moduli tenglamalar quyidagicha ekvivalent almashtirish bilan
yechiladi:
27
@matematika_variant
1. f ( x)  f ( x)  f ( x)  0 ; 2. f ( x)   f ( x)  f ( x)  0 ;
agar F ( x)  0,
 F ( x)  f ( x),
3. F ( x)  f ( x)   F ( x)   f ( x), agar F ( x)  0;

2
2
2
2
4. f ( x)  g ( x)  f ( x)  g ( x) ; 5. f ( x)  a (a  0)  f ( x)  a ;
agar x  a  0,
 F ( x, x  a)  0,
F
(
x
,
x

a
)

0

6.
 F ( x,  x  a)  0, agar x  a  0;

 f ( x)  g ( x),
f
(
x
)

g
(
x
)

7.
 f ( x)   g ( x);

 f ( x)  g ( x), agar x  0,
f
(
x
)

g
(
x
)

8.
 f ( x)  g ( x), agar x  0;

 f ( x)  a
f
(
x
)

a
(
a

0)

9.
 f ( x)  a ; 10. f ( x)  a (a  0)   .

Modulli tengsizliklar
Moduli tengsizliklar quyidagicha ekvivalent almashtirish bilan
yechiladi:
1. f ( x)  a (a  0)  a  f ( x)  a ;
2
2
2. f ( x)  a (a  0)  f ( x)  a yoki
 f ( x )  a,
f ( x)  a (a  0)  
agar a  0  x  R;
f
(
x
)


a
;

2
2
3. f ( x)   ( x)  f ( x)   ( x) ;
 f ( x)  g ( x), agar x  0,
f
(
x
)

g
(
x
)

 f ( x)  g ( x), agar x  0;
4.

 f ( x)  g ( x),
f
(
x
)

g
(
x
)

agar g ( x)  0  x ;

5.

f
(
x
)

g
(
x
);

 f ( x )  g ( x),
 f ( x)  g ( x) agar x  0,
6. f ( x )  g ( x)   f ( x )   g ( x) yoki  f ( x)  g ( x) agar x  0;


2n
2
7. a f ( x)  b f ( x)  c  0   0  f ( x)  y  a y  b y  c  0   0  ; y  0, n  N
n
n
Irrasional tenglama.
28
@matematika_variant
Irrasional
tenglamalarni
umumiy
holda
n  N  :
ekvivalent almashtirish yordamida yechish mumkin
1.
2n
 f ( x)  0,

f ( x)  2n  ( x)   ( x)  0,
2.
 f ( x)   ( x).

2n
quyidagicha
 f ( x)  0,

f ( x)   ( x)   ( x)  0,

2n
 f ( x)   ( x).
3. 2n f ( x)  a (a  0)  x .
4.
2n1
2n1
( x) .
f ( x)  2n1  ( x)  f ( x)   ( x) . 5. 2n1 f ( x)   ( x)  f ( x)  
f ( x)  0, (a  0),  ( x)  0,


f ( x)   ( x)  a  
2
f
(
x
)

a


(
x
)
.


 f ( x)  0,  ( x)  0, b   ( x)  0,

f ( x)   ( x)  b (b  0)  
2
f
(
x
)

b


(
x
)
.



6.


7.

Irrasional tengsizliklar
Irrasional tengsizliklar quyidagicha ekvivalent almashtirish
yordamida yechiladi  n  N  :
1.
3.
2n
 f ( x)  0,

f ( x)  g ( x)   g ( x)  0,
2.
 f ( x)  g 2 n ( x).

2n1
f ( x)  g ( x)  f ( x)  g 2n1( x).
4.
  g ( x)  0,

  f ( x)  0,
f ( x)  g ( x)  
 g ( x)  0,
 
  f ( x)  g 2n ( x).
2n
5.
6.
 g ( x)  0,

f ( x)
 1   f ( x)  0,
g ( x)

2n
 f ( x)  g ( x)
2n1
f ( x)  2n1 g ( x)  f ( x)  g ( x).
2n1
f ( x)  g ( x)  f ( x)  g 2n1( x).

 g ( x)  0,

2n

 f ( x)  g ( x).
2n
7.
2n
 f ( x)  0,

f ( x)  2n g ( x)   g ( x)  0,
 f ( x)  g ( x).

29
@matematika_variant
f ( x)
 g ( x)  0,
1 
g ( x)
 f ( x)  0
2n
8.

 g ( x)  0, f ( x)  0,

2n

 f ( x)  g ( x).
Arifmetik progressiya
1.
n
hadini topish formulasi: an  a1   n  1 d , n  N , bu yerda
d - ayirmasi, a1 - birinchi hadi, an n-chi hadi, n  hadlari soni.
2. d - ayirmani topish: d  a2  a1  a3  a2  a4  a3  ...  an  an1
an  am
d

yoki
nm
3. Xossalari: Chetki hadlar yig’indisi o’rta hadning ikkilnganiga teng
ank  an k
ak 1  ak 1
an 
a) ak 
yoki
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) Sn  a1  a2  a3  ...  an ; 2) Sn  Sn1  an ;
(a1  an )n
2a1  d (n  1)
 n ; Sn  n  a( n1) 2 ;
2
2
mn
S  Sn , m  n ;
4) Snk  Sn  Sk  n  k  d ; 5) Smn 
mn m
3) Sn 
;
Sn 


k
k
6) Sn  Sn  d  n  (k  1), Sn  n dan k gacha bo`lgan sonlar yig;
7) a2  a4  ...  a2 n  a1  a3  ...  a2 n1  n  d ;
Geometrik progressiya
n1
1. n  hadini topish formulasi: bn  b1q , n  N , bu yerda
q -maxraji, b1 - birinchi hadi, bn n-chi hadi, n  hadlari soni.
n 2
yoki bk  bk q nk ; bnk  bn q k , bnk  bn q k ;
a) bn  b2 q
n1
 b2 q n2  b3q n3  ...  bn2q 2  bn1q ;
b) bn  b1q
30
@matematika_variant
2. q -maxrajini topish:
b
b
b
b b
b
b
b
q  2  3  ...  n ; q 2  3  4 ; q3  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  bp agar m  m  k  p;
v) agar bk , bn , bm , bp ;
b
k , n, p  N bо`lsa,  k
 bn



k p
b
 k
 bp





k n
bо`ladi;
g) agar b1 , b2 , b3 , ..., b n , 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 (q n  1)
q 1

, Sn 
bn q  b1
q 1

2) Sn  Sn1  bn ;
, (q  1); S
4) Snm  2 Sn  Sm ; 5)
2n
toq b1 (q

2
n
 1)
q 1
,S
2n
juft b2 (q

2
n
b1  b3  ...  b2 n 1
b2  b4  ...  b2 n
 1)
q 1

;
1
q ;
k
n1 k
Sn  k chi haddan boshlab n ta hadi yig’indisi;
6) Sn  Sn q
7) geometrik progressiya hadlari soni toq bo`lsa,
b2  b4  ...  b2n   b1  b3  ...  b2n1  b2n1   b2 n1   q bо`ladi.
4. Agar geometrik progressiyada q  1 , q  0 bo`lsa, bu
progressiya cheksiz kamayuvchi geometrik progressiya deyiladi.
S - cheksiz kamayuvchi geometrik progressiya hadlari yig’indisi:
b1
b1
b2
juft
S
,  q  1 ; S toq 
,
S

.
1 q
1  q2
1  q2
5. Agar geometrik progressiyada q  1 , bo`lsa, bu progressiya
o`suvchi geometrik progressiya deyiladi.
31
@matematika_variant
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
M1  M 2 konsentrasiyasi z % : z %  1
bo’lgan
M1  M 2
eritma hosil bo’ladi.
32
@matematika_variant
FUNKSIYA
Aniqlanish sohasi (an.s.)
 ( x)
l.
y
2.
y  2 n f ( x) , n  N
bo’lsa, an.s.
f ( x)  0 bo’ladi.
3.
y  2n1 f ( x) ,
n N
bo’lsa, an.s.
  f ( x)   bo’ladi.
4.
y
n N
bo’lsa, an.s.
f ( x)  0 bo’ladi.
f ( x)
2n
bo’lsa, an.s. f ( x)  0 bo’ladi.
1
,
f ( x)
 f ( x)  0,

5. y  log g ( x) f ( x) bo’lsa, an.s.  g ( x)  0, bo’ladi.
 g ( x)  1;

6. y  arccos f ( x); y  arcsin f ( x) bo’lsa, an.s. 1  f ( x)  1
bo’ladi.

f
(
x
)

  n, n  Z bo’ladi.
y

tg
f
(
x
)
7.
bo’lsa, an.s.
2
8. y  ctg f ( x)
bo’lsa, an.s. f ( x)   n, n  Z bo’ladi.
9. y  arctg x
bo’lsa, an.s. x  R bo’ladi.
10. y  arcctg x
bo’lsa, an.s. x  R bo’ladi.
2
x
11. y  ax  bx  c; y  x ; y  a ; y  sin x; y  cos x bo’lsa,
an.s. x  R bo’ladi.
k
y

, k  R, k  0 bo’lsa, an.s. D( y)   ; 0   0;    bo’ladi.
12.
x
 f ( x)  0, g ( x)  0,

f ( x)  g ( x)
 ( x)  0,  ( x)  0,
13. y 
bo’lsa, an.s. 
bo’ladi.
 ( x)   ( x)

  ( x)   ( x)  0
Qiymatlar sohasi (q.s.)
x
1. y  a bo’lsa, q.s. E ( y)   0;    bo’ladi.
2. y  loga f ( x), a  0, a  1 bo’lsa, q.s. E ( y)   ;    bo’ladi.
33
@matematika_variant
2
2
2
2
3. y  a sin k x  bcos k x bo’lsa, q.s. E ( y)   a  b ; a  b  bo’ladi.
4. y  arccos x bo’lsa, q.s.
5. y  arcsin x bo’lsa, q.s.
6. y  arctg x
7. y  arcctg x
E ( y)   0;   bo’ladi.
  
E ( y )    ;  bo’ladi.
2 2


  
E
(
y
)

bo’lsa, q.s.
  ;  bo’ladi.
 2 2
bo’lsa, q.s. E ( y )   0;   bo’ladi.
 x0 , y0  :
2
8. y  ax  bx  c parabolaning uchi
4ac  b 2
y0 
,
4a
x0  
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 x0 ,
a) a  0 bo’lsa, q.s. E ( y)  


y ;  
b) a  0 bo’lsa, q.s. E ( y)  0;
y0 , y0  0 bo’lsa:
0
bo’ladi.
y0  bo’ladi.
10. y  x bo’lsa, q.s. E ( y)  0;    bo’ladi.
k
11. y  , k  R, k  0 bo’lsa, q.s. E ( y)   ; 0   0;    bo’ladi.
x
Funksiyaning juft va toqligi
1. f ( x)  f ( x) bo’lsa. funksiya juft.
2. f ( x)   f ( x) bo’lsa, funksiya toq.
3. Yuqoridagi ikkala tenglik ham bajarilmasa, funksiya juft ham,
toq ham emas.
4. y  x 2 , y  x 4 , y  x , y  cos x, y  a x  a  x funksiyalar juft.
3
2n1
 n  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  5x  2, y  x  3 , y  arccos x , y  arcctg x -
34
@matematika_variant
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) Juft  Toq  Juft ham, toq ham emas;
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 2 .
2. y  tgx, y  ctgx funksiyalarning eng kichik musbat (e.k.m.)
davri  .
2
3. y  sinkx, y  coskx funksiyalarning e.k.m. davri T 
.
4. y  tgkx, y  ctgkx funksiyalarning e.k.m. davri T1 

k
k
.
m
m
5. y  sin (ax  b), y  cos (ax  b) funksiyalarning e.k.m. davri
2

T

T

m  toq bo`lsa: 2
m  juft bo`lsa: 3
.
a teng;
a

T

6. y  tg (ax  b), y  c tg (ax  b) funksiyalarning e.k.m. davri 3
a.
m
m
6. Bir necha davriy funksiyalarning yig`indisidan iborat davriy
funksiyaning e.k.m. davrini topish uchun qo`shiluvchi
funksiyalar e.k.m. davrlarining EKUK ini olish kerak.
Masalan: y  7 cos(2 x  1)  3tg 0,5x  5sin 4 x funksiyalarning
e.k.m. davrini toping: T1 
2
2 

  , T2  2 . T3 

EKUK   , 2 ,   2 .
2
4 2
2

Chiziqli
funksiya
1. y  kx  b to’g’ri chiziq tenglamasi, bunda k  tg  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,
35
@matematika_variant
 b 
OX o’qini   ; 0  nuqtada kesib o`tadi.
 k 
3. y  k1 x  b1 va y  k2 x  b2 tenglama bilan berilgan to’g’ri
chiziqlar orasidagi  burchakni topish formulasi:
k k
tg  2 1 , k1k2  1 .
1  k1  k2
Xossalari:
a) k1  k2 ikki to’g’ri chiziqning parallellik sharti;
b) k1  k2  1 ikki to’g’ri chiziqning perpendikulyarlik sharti;
v) k1  k2 bo’lib, b1  b2 da to’g’ri chizilar ustma-ust tushadi;
g) k1  k2 bo’lib, b1  b2 da to’g’ri chizilar ustma-ust tushmaydi;
d) k1  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:
y3  y1
x  x1
 3
y2  y1
x2  x1 .
7. To’g’ri chiziqning umumiy ko’rinishdagi tenglamasi:
ax  by  c  0 , a, b, c  R .
8. M ( x0 , y0 ) nuqtadan ax  by  c  0 to’g’ri chiziqqacha masofa:
d
ax0  by0  c
a 2  b2
.
9. ax  by  c1  0 , ax  by  c2  0 parallel to’g’ri chiziqlar
36
@matematika_variant
orasidagi masofa:
d
c2  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


a) a
b2 c2 bo’lsa, parallel bo’ladi;
2
a1
b1
c1


b) a
b2 c2 bo’lsa, ustma-ust tushadi;
2
a1
b1

v) a
b2 bo’lsa, ular kesishadi.
2
11. To’g’ri chiziqning koordinata o’qlardan ajratgan kesmalarga
nisbatan tenglamasi:
x y
  1,
a b
c  a 2  b 2 -kesma uzunligi
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 .
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
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  2a  b .
simmetrik to’g’ri chiziq
16. y  kx  b to’g’ri chiziqqa y  x to’g’ri chiziqqa nisbatan
1
b
y

x

simmetrik to’g’ri chiziq
.
k
k
17. y  kx  b to’g’ri chiziqqa OY o’qiga nisbatan simmetrik
to’g’ri chiziq y  kx  b .
18. y  kx  b to’g’ri chiziqqa OX o’qiga nisbatan simmetrik
to’g’ri chiziq y  kx  b .
y  f (x) funksiyaga Ox ga nisbatan simmetrik
y   f (x)
y  f (  x)
Oy ga nisbatan simmetrik
Koordinata boshiga nisbatan simmetrik  y  f ( x)
37
@matematika_variant
Kvadratik funksiya
2
1. y  ax  bx  c , a  0 kvadratik funksiyaning umumiy ko’rinishi.
2
2. y  ax  bx  c , a  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 O X 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
4ac  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:
38
@matematika_variant
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.
Parabola tenglamasi:
2
1) y  ax  bx  c
2
2) y  a( x  x0 )  y0
3) y  a( x  x1 )( x  x2 )

Darajali funksiya y  x
l.
y  xn ,
n N :
D( y)  E ( y)   ;   .
D( y)   ;   , E ( y)  0;   ,
2.
y  xn  1 xn ,
n N :
39
@matematika_variant
D( y)   ;0    0;   , E( y)   0;   ,
n
3. y  x ,
n N :
D( y)  E ( y)   ;  
D( y)  E ( y)  0;  
p q
4. y  x ,
D( y)  E ( y)   ;0    0;   .
p, q  Z , q  0 :
D( y)  E ( y)   0;   .
D( y)  E ( y)  0;   ,
Grafiklarni o’zgartirish
40
@matematika_variant
Funksiyaning o’sishi va kamayishi
1. Agar x1 , x2   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   a ; b  bo`lib x1  x2 , f ( x1 )  f ( x2 ) bo`lsa, u holda
y  f ( x) kamayuvchi bo`ladi.
41
@matematika_variant
Ko’rsatkichli funksiyaning xossalari va grafigi
x
Ko’rsatkichli funksiyaning ko’rinishi: y  a
 a  0, a  1 .
1. Aniqlanish sohasi D( y)   ;    barcha haqiqiy sonlar
to’plami.
2. Qiymatlar sohasi E ( y)   0;    barcha musbat haqiqiy sonlar
to’plami.
3. Ko’rsatkichli funksiya a  1 bo’lganda barcha haqiqiy sonlar
to’plamida o’suvchi; agar 0  a  1 bo’lganda kamayuvchi.
4. Ko’rsatkichli funksiyaning grafigi (0; 1) nuqtadan o’tadi va OX
o’qidan yuqorida joylashgan.
5. 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, b  R  ko`rinishdagi tenglamalarga
sodda ko’rsatkichli tenglama diyiladi. Bundan:
 agar a  0, a  1, b  0 bo`lsa, teglama yechimga ega emas,
x
a

b


a)
log b
x
 agar a  0, a  1, b  0 bo`lsa, a  a a  x  log a b;
 1  a  0, a  1  f ( x)  0.
b) a
Yechishda qo’llaniladigan asosiy ekvivalent almashtirishlar:
f ( x)
 a ( x )  f ( x)   ( x), (a  0, a  1)
1. a
 agar f ( x)  0 bo`lsa, yechim yo ' q,
 ( x)
 f ( x) (a  0, a  1)  
2. a
 agar f ( x)  0 bo`lsa,  ( x)  log a f ( x).
f ( x)
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@matematika_variant
3.
f ( x)
g ( x)
 f ( x) quyidagi hollarda yechish mumkin:
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)  t  a , f (t )  0  a  t1, a  t2 , ..., a  tk .
f ( x)
   a f ( x)    a f ( x)  0   0,  ,   R; b2  ac  
5.   a
a
 
b
f ( x)
a
 t,  t   t    0   
b
f ( x)
2
a
 t1 ,  
b
f ( x)
 t2 .
f ( x)
   a f ( x)  c  0  ,  ,   R; a  b  1 
6.   a
 a f ( x)  t ,  t 2  ct    0  a f ( x)  t1 , a f ( x)  t2 .
7. 1  3
x2
 2  1 2  
x
x

3 2

x
x
x
f ( x)
 1  f ( x)  0.

  
 1   sin    cos   1  x  2.
6 
6

f ( x)
 b f ( x)  a, b  0; a, b  1   a b 
8. a
Ko’rsatkichli tengsizliklar
Ko’rsatkichli tengsizliklar ushbu ekvivalent almashtirish
yordamida yechiladi:
0  a  1,
a  1,
f ( x)
g ( x)
a

a



1.
 f ( x)  g ( x);
 f ( x)  g ( x).
2.
0  f ( x)  1,
 g ( x)  0;
 f ( x)  g ( x )  1  
f ( x)
3. a
 f ( x)  1,

 g ( x)  0.
 f ( x)  log a b, a  1, b  0,

 b   f ( x)  log a b, 0  a  1, b  0,
 x  D( f ), a  0, b  0.

f ( x)
b
4. a
 a  0, a  1,
b  0   yechimga ega emas.
LOGARIFM
logab  x  a x  b , a  1, a  0, b  0 .
loga b
Bundan asosiy logarifmik ayniyatni a
a  logarifmning asosi har doim a  1, a  0 .
Logarifmning xossalari
1)
log a a  1, a  1, a  0 ;
 b olamiz,
2) log a 1  0 ;
43
@matematika_variant
3)
loga ( X  Y )  loga X  loga Y , X  0, Y  0 ;
4)
log a b 
5)
7)
9)
1
; a, b  0; a, b  1 ;
logb a
X
log a    log a X  log a Y , X  0, Y  0 ; 6) log b p  p log b, p  R ;
a
a
Y 
1
p
log q b p  log a b, q  0, p, q  R ; 8) log q p  log a b ;
a
a
q
q
log a b 
logc b
logc a
, c  1, c  0 ;
11) loga b  logb c    log x y  log a y; 12) a
13)
15.
16.
17.
18.
19.
20.
21.
22.
23.
10)
a
log a b
log c
b
b
c
logb a
log a
b
;
, log a b  0 ;
loge x  ln x -natural logarifm; 14) log10 x  lg x  o'nli logarifm;
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  logb p bo`ladi ;
p  1, a  b  1 bo`lsa,
loga p  logb p bo`ladi ;
p  1, 0  a  b  1 bo`lsa,
loga p  logb p bo`ladi ;
0  p  1, 0  a  b  1 bo`lsa,
0  p  1, a  b  0 bo`lsa,
log a p  logb 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  loga x,  a  0, a  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.
44
@matematika_variant
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  log a x logarifmik funksiya juft ham, toq ham, davriy ham emas.
7. Logarifmik funksiyaning grafigi (1; 0) nuqtadan o’tadi.
8. y  loga x,  a  0, a  1, x  0  funksiyaning grafigi:
D( y)   0;    ,
E ( y)   ;    .
Logarifmik tenglamalar
Ushbu log a x  b  a  0, a  1, b  R  ko`rinishdagi
tenglamalarga sodda logarifmik tenglama diyiladi.
Yechishda qo’llaniladigan asosiy ekvivalent almashtirishlar:
b
1. log a x  b  x  a , x  0 (a  1, a  0) .
b
2. log a f ( x)  b  f ( x)  a , f ( x)  0, b  R (a  1, a  0) .

 f ( x)  0,  ( x)  0,  ( x)  1,
3. log ( x ) f ( x)  b  
b

 f ( x)   ( x).

 f ( x)  0, a  0, a  1,
log
f
(
x
)


(
x
)


4.
a
 ( x)
.

 f ( x)  a
 f ( x)  0, g ( x)  0, a  0, a  1,
5. log a f ( x)  log a g ( x)   f ( x)  g ( x).

 f ( x)  0,
 g ( x)  0,


6. log f ( x) A  log g ( x) A   f ( x)  1, A  0, yoki  g ( x)  1, A  0,
 f ( x)  g ( x);
 f ( x)  g ( x).


45
@matematika_variant
7.
f ( x)  a l
og a g ( x )
 f ( x)  0, g ( x)  0,

 a  0, a  1,
 f ( x)  g ( x).

 f ( x)  0, g ( x)  0,

 f ( x)  g ( x)  m  x  .

8. log a f ( x)  log a g ( x)  log a m( x)  a  0, a  1  
 g ( x)  0,
a

0,
a

1,
n

N


  2n1
( x)  g ( x).
 f
 f ( x)  0,
2
n
log
f
(
x
)

log
g
(
x
)
a

0,
a

1,
n

N


  2n
10.
a
a
 f ( x)  g ( x).
11. f (loga x)  0, a  0, a  1  log a x  t , f (t )  0.
9.  2n  1 log a f ( x)  log a g ( x)
12. loga x  logb x  logc x  d , a  0, b  0, c  0, a  1, b  1, c  1, x  0 
 log a x 
log a x
log a b

log a x
log a c
 d.
Logarifmik tengsizliklar
Logarifmik tengsizliklar ushbu ekvivalent almashtirish
yordamida yechiladi:
1. log a
3. log a
4.
5.
a  1,
0  a  1,


f ( x)  b   f ( x)  0, 2. log a f ( x)  b   f ( x)  0,
 f ( x)  a b .

b

 f ( x)  a .
0  a  1, g ( x)  0, a  1, g ( x)  0,


f ( x)  log a g ( x)   f ( x)  0,
 f ( x)  0,
 f ( x)  g ( x);
 f ( x)  g ( x).


0  f ( x)  1,


log f ( x ) g ( x)  a   g ( x)  0,

a

 g ( x)   f ( x)  ;
0  f ( x)  1,
log f ( x ) g ( x)  0  
0  g ( x)  1
0  f ( x)  1,
6. log f ( x ) g ( x)  0   g ( x)  1

 f ( x)  1,


 g ( x)  0,

a

 g ( x)   f ( x)  .
 f ( x)  1,

 g ( x)  1.
 f ( x)  1,

0  g ( x)  1.
46
@matematika_variant
0  f ( x)  1,
 f ( x)  1,
log
g
(
x
)

0



7.
f ( x)
0  g ( x)  1
 g ( x)  1.
0  f ( x)  1,
 f ( x)  1,

8. log f ( x ) g ( x)  0   g ( x)  1

0  g ( x)  1.
 ( x)  1,
 f ( x)  0,


0   ( x)  1,
9. log ( x ) f ( x)  log ( x ) g ( x)   g ( x)  0,
 f ( x)  g ( x);

 ( x)  1,
10.
log ( x ) f ( x)  log ( x )

g ( x)   f ( x)  0,
 f ( x)  g ( x);

 f ( x)  g ( x).

 g ( x)  0,

0   ( x)  1,
 f ( x)  g ( x).

47
@matematika_variant
TRIGONOMETRIYA
Boshlang’ich tushunchalar
0
1.  -gradusdan radianga o’tish:  rad 
2.  rad -radiandan gradusga o’tish: 


 .
180
180

  rad
.
3. Ta`riflar:
1) sin  
y
 y;
r
3) tg 
y
,
x
5) tg 
sin 
cos 
2) cos  
x  0 ; 4) ctg 
;
6) ctg 
x
 x;
r
x
,
y
y  0;
cos 
.
sin 
Trigonometrik funksiyalar qiymatlari jadvali
Burchak α,
gradus(radian)
0° (0)

15°  
 12 

18°  
 10 

22,5°  
8

30°  
8
sin α
0
3 1
2 2
Funksiyalar
cos α
tg α
1
0
3 1
2 3
2 2
5 1
ctg α
Mavjud emas
2 3
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
12
3 2
1

36°  
5 5
2 2
5 1
4
10  2 5
5 1
10  2 5
45° (π /4)
2 2
2 2
1
1
60° (π /3)
3 2
90° (π /2)
1
0
Mavjud emas
0
5
75°  
3 1
2 2
0
-1
0
3 1
2 2
2 3
2 3
-1
0
1
0
Mavjud emas
0
Mavjud emas
0
Mavjud emas
5
 12 
180° (π)
270° (3 π /2)
360° (2 π)
12
3
3
5 1
1
3
3
48
@matematika_variant
Trigonometrik funksiyalarning ishoralari
Asosiy trigonometrik ayniyatlar
sin 
1


;    2n  1 , n  Z .
cos  ctg
2
1. cos2   sin 2   1 .
2. tg 
3. tg  ctg  1 .
4. ctg 
2
5. 1  tg  
1
.
cos 2 
cos 
1

;    n, n  Z .
sin  tg
1
2
1

ctg


;    n, n  Z .
6.
sin 2 
Trigonometrik funksiyalarning birini ikkinchisi orqali
ifodalash
2
1. cos    1  sin   
ctg

1
.
1  ctg 2
1  tg 2
tg
1
2
sin



1

cos





2.
.
1  tg 2
1  ctg 2
1
1  cos 2 
sin 
tg






3.
.
ctg
cos 
1  sin 2 
1
1  sin 2 
cos 


4. ctg 
.
tg
sin 
1  cos 2 
 - ning qaysi chorakka tegishliligiga qarab "+" yoki "-" ishoradan
 


 0;  , ya`ni I-chorakda bo`lsa,
biri olinadi. Masalan: Agar
2


 3 


  ;  , ya`ni III-chorakda
1- formulada  olinadi; agar
2 

bo`lsa, 1- formulada  olinadi.
49
@matematika_variant
1. cos (
2. cos (
3. sin (
4. sin (
 )
 )
 )
 )
Qo'shish formulalari
 cos  cos   sin  sin  .
 cos  cos   sin  sin  .
 sin  cos   cos  sin  .
 sin  cos   cos  sin  .
tg  tg 
ctg  ctg  1
5. tg (   ) 
6. ctg (   ) 
.
.
1 tg  tg 
ctg  ctg 
Karrali burchaklar (Ikkilanagan va uchlangan burchak)
2. cos 2  cos 2  sin2
1. sin 2  2sin  cos  .
2tg
3. sin 2 
.
1  tg 2
1  tg 2
4. cos 2 
.
1  tg 2
5. cos 2  2cos   1  1  2sin  . 6. tg 2 
2
2
ctg 2  1
7. ctg 2 
.
2ctg
9. sin 3  3sin  - 4sin3 .
8. tg 2 
3ctg  ctg 3
12. ctg 3 
.
1  3ctg 2
14. cos 4  8cos 4  8cos 2  1.
13. sin 4  cos    4sin  - 8sin3  .
15. tg 4 
1  6tg 2  tg 4
2
.
ctg  tg
10. cos 3  4cos3 - 3cos 
3tg  tg 3
11. tg 3 
.
1  3tg 2
4tg  1  tg 2 
2tg
.
1  tg 2
ctg 4  6ctg 2  1
16. ctg 4 
.
4ctg 2   ctg 2  1
.
Darajasini pasaytirish
1  cos 2
.
2
3sin   cos 3
3
sin


3.
.
4
1  cos 2
2
tg


5.
1  cos 2 .
2
1. sin  
4
4
7. sin   cos   cos 2 .
1
8
4
9. sin    cos 4  4cos 2  3 .
2
2. cos  
4.
cos 3 
1  cos 2
.
2
3cos   cos 3
4
.
1  cos 2
2
ctg


6.
1  cos 2 .
4
4
8. cos   sin   cos 2 .
4
10. cos  
1
 cos 4  4cos 2  3 .
8
50
@matematika_variant
4
4
11. cos   sin  
3 1
 cos 4 .
4 4
6
6
12. cos   sin  
5 3
 cos 4 .
8 8
Yarim burchak uchun formulalar

 sin2
1)
cos 2
3)
cos   cos 2
5)
cos
7)
ctg
2


2


2
4) sin 2  

2
cos

2
;
1  cos 
;
2

1  cos 
tg


6)
2
1  cos  ;
;
1  cos 
2 
cos

8)
;
2
2
1  cos 
2 
tg

10)
2 1  cos  ;
1  cos 
1  cos  ;
1  cos 
9)
;
2
2
1  cos 
2 


ctg

tg 2  ctg 2  1 ;
11)
;
12)
2 1  cos 
2
2
 1  cos 
sin 
 1  cos 
sin 
ctg


tg


13)
2
sin 
1  cos  ; 14)
2
sin 
1  cos  .
sin2

2
2) sin   2sin
 1;


 sin2 ;
2
2
1  cos 

2


Trigonometrik funksiyalarni yarim burchak tangensi
orqali ifodasi

2  

2  
cos


1

tg
1

tg
;
2)




2
2 
2

 

2  
2 
3) tg  2tg
1  tg
 ; 4) ctg  1  tg
2 
2
2

1) sin   2tg

2  
1

tg

;
2





2tg

2
.
Ko'paytmani yig'indiga keltirish
1
 sin      sin      .
2
1
cos


cos


cos      cos     .
2.
2
1. sin   cos  
51
@matematika_variant
1
cos      cos     .
2
tg  tg 
tg   tg

4. tg  tg  
.
ctg  ctg  ctg   ctg
ctg  ctg  ctg   ctg
ctg


ctg



5.
.
tg  tg 
tg  tg 
tg  ctg  tg  ctg 
tg


ctg



6.
ctg  tg  tg   ctg .
3. sin   sin  
7. cos   cos 2  cos 4  ...  cos 2  
n
sin 2n1
.
2n1 sin 

2
3
n
1
cos

cos

cos

...

cos

8.
2n  1
2n  1
2n  1
2n  1 2n .

2
1
 2 , 5  2  2  1, n  2 ;
Masalan: a) cos  cos
5
5
2

2
7
1
b)
9.
cos
cos   cos

15
2
 cos
 cos
15

4
 ...  cos
 ...  cos
15

2n


27
, 15  2  7  1, n  7 .
sin 2
2n1 sin
 .
2n
Yig’indini ko’paytmaga keltirish
 
 
 cos
. 2.
2
2
 
 
cos


cos


2
cos

cos
3.
. 4.
2
2
sin (   )
tg


tg


5.
6.
cos   cos  .
sin (   )
ctg


ctg


7.
8.
sin   sin  .
1. sin   sin   2sin
 
sin
x

cos
x

2
sin
x .
9.
4

2
11. 1  cos   2sin

2
.
sin   sin   2sin
 
 
 cos
2
2 .
   
cos   cos   2sin
 sin
2
2 .
sin (   )
tg  tg  
cos   cos  .
sin (    )
ctg  ctg  
sin   sin  .
2
10. 1  cos   2cos

2
;
 
12. sin x  3cos x  2sin  x  3  .


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@matematika_variant
 
sin
x

cos
x

2
sin
 x  .
13.
4


15. 3 sin x  cos x  2sin  x 



sin
x

3
cos
x

2
sin
x

14.

.
3


.
6
kx
(k  1) x
x
 sin
sin .
2
2
2
kx
(k  1) x
x
17. cos x  cos 2 x  cos 3x  ...  cos kx  sin  cos
sin .
2
2
2
16. sin x  sin 2 x  sin 3x  ...  sin kx  sin
18. sin   sin 3  sin 5  ...  sin (2n 1)  sin2 n sin  .
19. cos   cos 3  cos 5  ...  cos (2n 1)  sin n  cos n sin  .
nk
n  k 1
x
x  cos
x sin .
2
2
2

3
5
(2n  1)

21. sin  sin
 sin
 ...  sin
 1  cos n  2sin .
20. cos kx  cos (k  1) x  ...  cos nx  cos
2
2
2
2
2
Muhim trigonometrik shakl almashtirishlar
6.
sin 3
.
4
cos 3
cos   cos(60   )  cos (60   ) 
.
4
tg 3
tg  tg (60   )  tg (60   ) 
.
4
ctg 3
ctg  ctg (60   )  ctg (60   ) 
.
4
sin 8
cos   cos 2  cos 4 
8sin  .
sin16
cos   cos 2  cos 4  cos 8 
16sin  .
7.
 sin x  cos x 
1. sin   sin (60   )  sin (60   ) 
2.
3.
4.
5.
2
 1  sin 2 x .
4
4
2
2
8. cos x  sin x  cos x  sin x  cos 2x .
1  cos 2 2 x
sin2 2 x 3  cos 4 x
 1

9. cos   sin  
.
2
2
4
1
1
6
6
cos


sin


5

3
cos
4
x

1  3cos 2 2 x .


10.
8
4
4
4


53
@matematika_variant
1
8
8
cos


sin


cos 2 x  3  cos 4 x  .
11.
4
Keltirish formulalari
Γ
3
3


    2   2  
        
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
sinα
cosα
tgα
ctgα
TRIGONOMETRIK 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  2 , ya'ni
sin( x  2 )  sinx, x  R.
4. y  sinx funksiya toq, ya'ni sin( x)  sinx.
 


2
2

5. Funksiya    2 n;  2 n  , n  Z kesmalarda 1 dan 1
2
 2

gacha o'sadi.
3


 2 n  , n  Z kesmalarda 1 dan 1
6. Funksiya   2 n;
gacha kamayadi.
7. Funksiyaning nollari: sinx  0  x   n, n  Z .
8. y  sinx funksiya x   2  2 n, n  Z nuqtalarda eng katta
qiymatga erishadi va u 1 ga teng.
9. y  sinx funksiya x  3 2  2 n, n  Z nuqtalarda eng kichik
qiymatga erishadi va u 1 ga teng.
10. Musbat qiymatlami: sinx  0  x   2 n;   2 n  , n  Z .
54
@matematika_variant
11. Manfiy qiymatlami: sinx  0  x    2 n; 2  2 n  , n  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  2 , ya'ni
cos( x  2 )  cosx, x  R.
4. y  cosx funksiya juft, ya'ni cos( x)  cosx .
5. Funksiya    2 n; 2 n , n  Z kesmalarda 1 dan 1
gacha o`sadi.
6. Funksiya  2 n;   2 n , n  Z kesmalarda 1 dan 1
gacha kamayadi.
7. Funksiyaning nollari: cosx  0  x   2   n, n  Z .
y  cosx funksiya x  2 n, n  Z nuqtalarda eng katta
qiymatga erishadi va u 1 ga teng.
9. y  cosx funksiya x    2 n, n  Z nuqtalarda eng kichik
8.
qiymatga erishadi va u 1 ga teng.
10. Musbat qiymatlami:
cosx  0  x     2  2 n;  2  2 n  , n  Z .
11. Manfiy qiymatlami:
cosx  0  x   2  2 n; 3 2  2 n  , n  Z .
55
@matematika_variant
12. y  cosx funksiyaning grafigi:
y  tgx funksiyaning xossalari va grafigi
1. Aniqlanish sohasi: x   2   n, n  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   )  tgx, x  D  tg  .
T   , ya'ni
4. y  tgx funksiya toq, ya'ni tg ( x)  tgx, x  D  tg  .
5. Funksiyaning nollari: tgx  0  x   n, n  Z .
6. Musbat qiymatlami: tgx  0  x   n;  2   n  , n  Z .
7. Manfiy qiymatlami: tgx  0  x     2   n;  n  , n  Z .
8. y  tgx funksiya    2  2 n;  2  2 n  , n  Z oraliqlarda o'sadi.
9. y  tgx funksiyaning grafigi:
56
@matematika_variant
y  ctgx funksiyaning xossalari va grafigi
1. Aniqlanish sohasi: x   n, n  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   , ya'ni
ctg ( x   )  ctgx, x  D  ctg  .
4. y  ctgx funksiya toq, ya'ni ctg ( x)  ctgx, x  D  ctg  .
5. Funksiyaning nollari: ctgx  0  x   2   n, n  Z .
6. Musbat qiymatlami: ctgx  0  x   n;  2   n  , n  Z .
7. Manfiy qiymatlami: ctgx  0  x     2   n;  n  , n  Z .
8. y  ctgx funksiya  n;    n  , n  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  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
57
@matematika_variant
x  2 aniqlanish sohasi. 1) y  4 
2) x  y  y 
Demak, y 
5
2;
x4
5
5
 x2
 2;
x2
y4
3) D( y)   ;4    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.
  
2. Aniqlanish sohasi: D( y)   1;1 . 3.Qiymatlar sohasi: E ( y)   ;  .
 2 2
4. Funksiya toq, ya'ni arcsin( x)  arcsinx .
5. Arksinusning ba`zi qiymatlari:
x
0
1
2
arcsinx
0

6
2
2

4
3
2

3
1


2

1
2

6
2

2
3

2
-1




4

3

2
6. y  arcsinx funksiya grafigi:
a) sin(arcsinx)  x, agar x   1;1;
b) arcsinx(sinx)  x,
c)


2
 arcsinx 
  
agar x   ;  ;
 2 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;   .
58
@matematika_variant
4. Funksiya juft ham, toq ham emas.
5. arccos ( x)    arccos x .
6. Arkkosinusning ba`zi qiymatlari:
x
0
1
2
arc cos x 

3
2
2
2

4
3
2

6
1

0
2
3
1
2

2
2

3
4
3
2
5
6
-1

7. y  arccosx funksiya grafigi:
a) cos(arccosx)  x, agar x   1;1;
agar x  0;  ;
b) arccos(cosx)  x,
c) 0  arccosx   .
ARKTANGENS
1. y  arctgx funksiya  ; +  oraliqda o'suvchi va bir qiymatli
aniqlangan.
2. Aniqlanish sohasi: D( y)   ;   .
3. Qiymatlar sohasi: E ( y)   0,5 ; 0,5  .
4. Funksiya toq, ya'ni arctg ( x)  arctgx .
5. Arktangensning ba`zi qiymatlari:
x
0
1
3
1
arctgx
0

6

4
3

3


1
3

6
-1


4
 3


3
6. y  arctgx funksiya grafigi:
a) tg (arctgx)  x, agar x   ;   ;
b)
c)
arctg (tgx)  x,


2
agar
 arctgx 

2
  
x   ; ;
 2 2
.
59
@matematika_variant
ARKKOTANGENS
1. y  arcctgx funksiya   ;+  oraliqda kamayuvchi va bir qiymaili
aniqlangan.
2. Aniqlanish sohasi: D( y)   ;   .
3. Qiymatlar sohasi: E ( y)   0;   .
4. Funksiya juft ham, toq ham emas. arcctg ( x)    arcctgx .
5. Arkkotangensning ba`zi qiymatlari:
0
1
-1  3
1
3  1
3
3
arcctgx 
2
3
5



3
4
6
2
3
4
6
6. y  arctgx funksiya grafigi:
a) ctg (arcctgx)  x, agar x   ;   ;
agar x   0;   ;
b) arcctg (ctgx)  x,
c) 0  arcctgx   .
Teskari trigonometrik funksiyalar ustida amallar
1. arcsin x  arccos x 


2. arctgx  arcctgx  .
.
2
2
2
3. sin(arccos x)   1  x , x  1. 4. cos(arcsin x)   1  x 2 , x  1.
1
1
ctg
arctgx

, x  0.
tg
arcctgx

,
x

0




5.
.
6.
x
x
7.
tg  arcsin x  
9. sin  arctg x  
11. cos  arctg x  
x
 1  x2
x
 1 x
1
 1 x
2
2
2
x  1. 8. tg  arccos x    1  x ,
,
x  1.
x
.
10. sin  arcctg x  
.
12. cos  arcctg x  
1
 1 x
x
.
2
 1 x
2
.
60
@matematika_variant




arccos xy  1  x 2  1  y 2 , x  y,

13. arccos x  arccos y  
 arccos xy  1  x 2  1  y 2 , x  y.

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  y.
, x   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.
x  1.
21.
2
22. cos(2arccos x)  2 x 1, x  1.
2x
, x  1.
23. tg (2arctg x) 
1  x2
25.
sin(2arccos x)  2 x 1  x 2 ,
2
22. cos(2arc s in x)  1  2 x ,
24. sin(2arctg x) 
1  x2
cos(2arctg x) 
,    x  .
1  x2
2x
,    x  .
1  x2
26. sin(2arcctg x) 
2
2
27. cos(2arcctg x)   1  x  1  x  ,    x  .
x  1.
2x
,    x  .
1  x2
Trigonometrik tenglamalar
1. sinx  a, a  1  x   1 arcsina   n, n  Z .
Xususiy hollar:
a) sinx  0  x   n, n  Z ;
b) sinx  1,
 x   2  2 n,
n  Z;
v) sinx  1,  x    2  2 n, n  Z ;
n
g ) sin2 x  a, 0  a  1  x  arcsin a   n, n  Z .
2. cosx  a, a  1  x  arccosa  2 n, n  Z .
Xususiy hollar:
a) cosx  0
 x   2   n, n  Z ;
b) cosx  1,
 x  2 n,
n  Z;
v) cosx  1,  x    2 n, n  Z ;
g ) cos 2 x  a, 0  a  1  x  arccos a   n, n  Z .
3. tgx  a, a  R  x  arctga   n, n  Z .
Xususiy hollar:
a) tgx  0  x   n, n  Z ;
61
@matematika_variant
 x    4   n,
b) tgx  1,
n  Z;
v) tg 2 x  a, 0  a    x  arctg a   n, n  Z .
4. ctgx  a, a  R  x  arcctga   n, n  Z .
Xususiy hollar:
a) ctgx  0  x   2   n, n  Z ;
b) ctgx  1,
 x    4   n,
n  Z;
v) ctg 2 x  a, 0  a    x  arcctg a   n, n  Z .
a
b
c
a

sinx

bcosx

c

sinx

cosx


5.
2
2
2
2
2
2
a b
a b
a b
 sinx  cos  cosx  sin 
bunda
cos  a
c
a b
2
a 2  b2 ,
2
 sin( x   ) 
sin  b
c
a b
2
a 2  b2 ,
2
,
c
a b
2
2
 1,
tg  b a .
 ax  b   cx  d   2 n,
sin
(
ax

b
)

sin
(
cx

d
)


6.
 ax  b  cx  d   2n  1  , n  Z .
 ax  b   cx  d   2 n,
7. cos(ax  b)  cos (cx  d )   ax  b  cx  d  2 n, n  Z .

ax  b  cx  d    n, n  Z ,

tg (ax  b)  tg (cx  d )  


8.
ax

b



n
,
cx

d

  n.


2
2
ax  b  cx  d    n,
ctg
(
ax

b
)


ctg
(
cx

d
)


9.
ax  b   n, cx  d   n, n  Z .
Trigonometrik tengsizliklar
1. sinx  a, a  1  x   arcsina  2 n;  arcsina  2 n  , n  Z .
2. sinx  a, a  1  x   arcsina  2 n;  arcsina  2 n , n  Z .
3. sinx  a, a  1  x    arcsina  2 n; arcsina  2 n , n  Z .
4. cosx  a, a  1  x   arccosa  2 n; arccosa  2 n , n  Z .
5. cosx  a, a  1  x   arccosa  2 n;  arccosa  2 (n  1 , n  Z.
6. tgx  a, a  R

x   arctga   n;  2   n  , n  Z .
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@matematika_variant
7. tgx  a, a  R

8. ctgx  a, a  R
9.
10
12.
13.
x     2   n; arctga   n , n  Z .
x   arcctga   n;    n  , n  Z .

ctgx  a, a  R  x   n; arcctga   n , n  Z .
arctgx  arctgy  x  y.
11. arcctgx  arcctgy  x  y.
arcsinx  arcsiny  1  y  x  1.
arccosx  arccosy  1  x  y  1.
Kvadratik, ko`rsatkchli, logarifmik, trigonomеtrik
funksiyalari o`zining aniqlanish sohasida uzluksiz.
FUNKSIYANING LIMITI
Agar ixtiyoriy   0 son uchun shunday   0 son topilsaki,
argument x ning 0  x  a   tengsizlikni qanoatlantiruvchi barcha
qiymatlarida
f ( x)  b  
funksiyaning
a
nuqtadagi
quyidagicha yoziladi:
b
tengsizlik bajarilsa,
x  a
son
f ( x)
dagi  limiti deb ataladi
va
lim f ( x)  b.
xa
1. Limitning xossalari: Agar lim f ( x)  A va
xa
lim g ( x)  B
xa
limitlar mavjud bo`lsa, u holda:
lim  f ( x)  g ( x)   lim f ( x)  lim g ( x)  A  B;
a)
b)
xa
xa
lim  f ( x)  g ( x)  lim f ( x)  lim g ( x)  A  B;
xa
xa
xa
lim  f ( x) g ( x)   lim f ( x)
v)
g)
xa
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
lim g ( x)  A B , B  0;
xa
bo`ladi.
n
 1
6. lim 1    e  2,71183... .
n 
n
1
x) x
sin px
px
 e.
 lim
 p, p  R . 7. lim (1 
x0
x0 sin x
x0
x
tg x
x
x
 lim
 1.
3. lim
8. lim x  1 .
x0 x
x0 tg x
x0
2. lim
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@matematika_variant
ax 1
 ln a, a  0 .
4. lim
x0 x
9.
ln  x  1
lim
 1.
5.
x0
x

x  1  1

 ,   0 .
10. lim
x0
arcsin x
x
 lim
 1.
x0
x0 arcsin x
x
lim
x
HOSILA
1. x va
x0  erkli o`zgaruvchilar y  f ( x) funksiyaning aniqlanish
sohasidan olingan qiymatlar bo`lsin, x  x  x0 ayirma erkli
o`zgaruvchining x0 nuqtadagi orttirmasi deyiladi.
Bundan x  x0  x .
2. y  f ( x0 )  f ( x0  x)  f ( x0 ) ga y  f ( x) funksiyaning x0
nuqtadagi orttirmasi deyiladi. Bundan f ( x0  x)  f ( x0 )  f ( x0 ) .
3. y  f ( x) funksiyaning x0 nuqtadagi hosilasi:
f ( x0 )
f ( x0  x)  f ( x0 )
y  lim
 lim
 f ( x0 ).
x 0
x 0
x
x
4. Hosilaning fizik va mexanik ma`nosi. Moddiy nuqta S  S  t 
qonuniyat bilan harakatlanayotgan bo`lsa, u holda:
a) S (t )   (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  bo`lsa, u
holda: a) k  f ( x0 ); b) tg  f ( x0 );
v) y  f ( x) funksiyaga x  x0
nuqtada o`tqazilgan urinma tenglamasi:
y  f ( x0 )  f ( x0 )  x  x0  .
6.
 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.
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@matematika_variant
8. y  f ( x) va y  g ( x) funksiyalarga M ( x0 , y0 ) nuqtada
o`tqazilgan urinmalar orasidagi burchakni topish:
g ( x0 )  f ( x0 )
, agar
1  f ( x0 )  g ( x0 )
a)
tg 
b)
  900 ,
agar
1  f ( x0 )  g ( x0 )  0;
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:
 y1  y0  f ( x0 )  x1  x0  ,

 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)  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
 1 
   2.
x
 x
5.
9. loga x  =


6.
 ex   ex .
7.
 x    x 1.
 a x   a xln a.
4.
8.
 x   2 1 x .
 ln x  
1
.
x
1
1
. 10.  sin x   cos x. 11.  cos x   sin x. 12.  tg x   2 .
xln a
cos x
13.  ctg x   
1
1
1


.
14.
arcsin
x

.
15.
arccos
x


.




sin2 x
1  x2
1  x2
1
1
16.  arctg x  
.
17.  arcctg x   
.
2
1 x
1  x2
Hosilalarni hisoblash qoidalari
Agar u  u ( x) va    ( x) bo'lsa, u holda:
1) ayirma va yig'indining hosilasi: u    u  ;
2) agar c  const bo'lsa,
 c  u 
 c  u ;
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@matematika_variant
3) ko'paytmaning hosilasi:
 u    u   u   ;
 u  u     u   
4) bo'linmaning hosilasi:   
.
2
 
Murakkab funksiyaning hosilasi
1.

f ( x)

3.
 e   e
5.
 lnf ( x)  
7.
 sinf ( x)   cos f ( x)  f ( x).
f ( x)
f ( x)
f ( x).
f ( x)
.
f ( x)
f ( x)
.
cos 2 f ( x)
f ( x)
11.  arcsinf ( x)  
.
2
1  f ( x)
9.
 tg f ( x)  
13.  arctg f ( x)  
15.
17.
 1 
f ( x)
2. 
.
  2
f ( x)
 f ( x) 
f ( x)

.
2 f ( x)

1
 a   a
6.
 loga f ( x)  =
8.
 cos f ( x)   sin f ( x)  f ( x).
f ( x)
f ( x)
 lna  f ( x).
f ( x)
.
f ( x)lna
f ( x)
.
sin2 f ( x)
f ( x)
12.  arccosf ( x)   
.
2
1  f ( x)
10.
f ( x)
.
2
1  f ( x)
 f  ( x)   f 
4.
 ctg f ( x)   
f ( x)
.
2
1  f ( x)
f ( x)

f ( x) 
.
n n1
n  f ( x)
14.  arcctg f ( x)   
( x) f ( x).
16.
n

 1 
f ( x)
.
 n
  
n 1
n
f
(
x
)
n f
( x)


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.
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@matematika_variant
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).
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:
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@matematika_variant
a) f ( x)  0  xi   a, b yoki xi   a, b , i  1, 2,3,... aniqlash;
b) agar xi   a, b bo`lsa, f ( x1 ), f ( x2 ), f ( x2 ),..., f (a), f (b)
ni hisoblash;
v) agar xi   a, b bo`lsa, f (a), f (b) ni hisoblash;
g) bu qiymatlar ichidan eng kattasi va eng kichigi tanlab olinadi.
2. y  sink x va y  cos k x funksiyalar uchun
max y  1, min y  1.
3. y  asinkx  bcoskx funksiya uchun esa
max y  a 2  b2 , min y   a 2  b2 .
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)  F ( x)  C deb
boshlang'ich funksiyasi deyiladi va
belgilanadi, C  ixtiyory o`zgarmas son.
Funksiyaning
1. C  Cx  C0 . 2. (kx  b)n 
boshlang'ichlari
(kx  b)n1
1
 C  n  1 . 3. ekxb  ekxb  C.
k (n  1)
k
1
1
 ln x  C.
5. sin (kx  b)   cos (kx  b)  C.
x
k
1
1
6. cos (kx  b)  sin (kx  b)  C.
7. tg (kx  b)   ln cos (kx  b)  C.
k
k
1
1
1
kx  b
8. ctg (kx  b)  ln sin(kx  b)  C.
9.
 ln tg
 C.
k
sin(kx  b)
k
2
1
1
1
1
 kx  b  
4.
10.
cos(kx  b)
 ln tg 
k

2

  ctg (kx  b)  C.
  C. 11.
2
sin2 (kx  b)
k
12.
1
1
1

tg
(
kx

b
)

C
.
13.
cos 2 (kx  b)
k
x2  a2
14.
1
1
x

arctg
 C.
x2  a2
a
a
16.
1
x a
2
 ln x  x  a  C. 17. a
2
2
15.
2
k x b

1
a2  x2
1
xa
ln
+C.
2a x  a
 arcsin
x
 C.
a
ak x  b

 C , a  0, a  1.
k  lna
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@matematika_variant
18. a  bx 
19. x  a
2
2
2
3b
 a  bx 
3
 C.
x
a2
2
2
  x  a  ln x  x 2  a 2  C.
2
2
INTEGRALLAR
b
1. N'yuton-Leybnis formulasi:
S   f ( x)dx  F ( x) ba  F (b)  F (a).
a
2. Egri chiziq bilan chegaralangan yuzalarni hisoblash:
b
a) Egrichiziqli trapesiya yuzi: S   f ( x)dx ;
a
b) agar f1 ( x)  f 2 ( x)  0 bo`lsa, u holda
b
S
  f ( x) 
1
f 2 ( x) dx; bo`ladi.
a
3. y  f ( x)  f ( x)  0  egri chiziq
aylanganda hosil bo'lgan jism hajmi:
b
b
V    f ( x)dx    y 2 dx.
2
a
a
b
4. AB : y  f ( x), a  x  b yoyning uzunligi:
l
1  f 2 ( x)dx .
a
 x  x(t ),
5. AB :  y  y(t ),   t   yoyning uzunligi:


l
x2 (t )  y2 (t )dt .

6. y  f ( x)  f ( x)  0  , x   a, b egri chiziqni OX o`qi atrofida
aylantirishdan hosil bo'lgan aylanish sirtining yuzini topish:
b
S  2 
f ( x)  1  f 2 ( x)dx .
0
Integrallash qoidasi
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b
1.
k
b
f ( x) dx  k  f ( x ) dx, k  const.
a
b
2.
a
b
  f ( x)  g ( x) dx  
a
b
3.
4.

f ( x)dx   g ( x )dx.
a
 f ( x) d g ( x) 
a
b
b
a
b
f ( x)  g ( x)
b
a
  g ( x) d f ( x).
a
b
f (kx  c)dx  1 k  f (kx  c) a , k  0, c  o`zgarmas sonlar.
a
5. Agar f ( x)  f ( x), x   a; a , a  0 bo`lsa,
a

a
a
f ( x)dx  2 f ( x)dx
0
bo`ladi.
6. Agar f ( x)   f ( x), x   a; a  , a  0 bo`lsa,
7. Agar f ( x)  0, x   a; b bo`lsa,
a
 f ( x)dx  0 bo`ladi.
a
b
 f ( x)dx  0 bo`ladi.
a
8. Agar a  x  c da f ( x)  0 ; c  x  b da f ( x)  0 bo`lsa,
b

c
a
a
c
f ( x) dx   f ( x)dx   f ( x)dx bo`ladi.
a
Aniqmas integral
1.
3. 
dx
 x  lnx  ln ln x  C.
x dx
1  x2
  1  x 2  C. 4.
2.
m
 sin x  cosxdx 
1
sinm 1 x  C.
n 1
 ln x
1 

 1
x

ln
x
dx

x



  C   1 .

   1   12 


x 2
a2
x
2
5.  a  x dx 
a  x  arcsin  C.
2
2
a
1
6.  arctgx dx  x  arctgx   ln 1  x 2   C.
2
7.  x  e x dx   x  1  e x  C. 8.  x 2e x dx   x 2  2 x  2   e x  C.
2
2
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x 1
 sin2 x  C.
2 4
cos3 x
3
11.  sin x dx  cos x 
 C.
3
9.
x 1
 sin2 x  C.
2 4
sin3 x
3
12.  cos x dx  sinx 
 C.
3
2
 sin xdx 
10.
2
 cos xdx 
13.


 1
ln
xdx

x

ln
x


ln
x dx  C.


14.
2
arcsin
x
dx

x

arcsin
x

1

x
 C.

2
2cx  b

arctg
 C , agar b 2  4ac,

2
2
4ac  b
dx
 4ac  b
15. 

a  bx  cx 2 
1
2cx  b  b 2  4ac
ln
 C , agar b 2  4ac.
 b 2  4ac 2cx  b  b 2  4ac

dx
1
16. 

ln 2cx  b  2 c a  bx  cx 2  C , c  0.
c
a  bx  cx 2
17.

a  bx  cx 2 dx 

b 2  4ac
8 c2
ln 2cx  b  2 c a  bx  cx 2  C.
dx
18. 

a  bx  cx 2
19.

20.

21.

22.

23.

2cx  b
a  bx  cx 2 
4c
1
b  4ac
2
ln
2cx  b  b 2  4ac
2cx  b  b  4ac
2
 C.
1
2cx  b
arcsin
 C , c  0.
2
2
c
a  bx  cx
b  4ac
2cx  b
a  bx  cx 2 dx 
a  bx  cx 2 
4c
2
b  4ac
2cx  b

arcsin
 C.
8 c2
b2  4ac
ax
dx   a  x  b  x    a  b  ln a  x  b  x  C.
b x
ax
ax
dx   a  x  b  x    a  b  arcsin
 C.
b x
ab
ax
bx
dx    a  x  b  x    a  b  arcsin
 C.
bx
ab
dx



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24.
 shxdx  chx  C,
 chxdx  shx  C.
25.
 thxdx  lnchx  C,
26.
 sin mx  sin nx dx  
27.
 cos mx  cos nx dx 
28.
 sin mx  cos nx dx  
29.
e
e
30.
 cthxdx  lnshx  C.
sin  m  n  x sin  m  n  x

 C , m  n.
2  m  n
2  m  n
sin  m  n  x sin  m  n  x

 C , m  n.
2  m  n
2 m  n
cos  m  n  x cos  m  n  x

 C , m  n.
2  m  n
2  m  n
ax
 sin nx dx  eax  a  sin nx  n  cos nx  (a 2  n 2 )  C.
ax
 cos nx dx  eax  a  sin nx  n  cos nx  (a 2  n 2 )  C.



dx

31. 
 
a  bcosx








dx
32. 
 
a  b s inx




 a b
x
arctg 
 tg   C , agar a  b,
2
a 2  b2
 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  b 2  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 .
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@matematika_variant
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   Cn0an  Cn1an1x  Cn2a n2 x2  ...  Cnk a nk xk  ...  Cnn xn
Q O’ 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
n  n   n 
nol bilan tugashi: x      2    3   ... .
 5  5  5 
2.   soat va minut strelkalari orasidagi burchak,   t vaqtdan
2
0
keyin ular orasidagi burchak bo`lsa, t   360  (   )  bo`ladi.
11
3. lg 2  0,3010, lg 3  0, 4771,
lg 300  lg8  lg 3  lg102  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   a, b  nuqtada f ( x0 ), ( j  1, 2, ..., n  1)
hosilalar mavjud bo`lsa, u holda
f ( x0 )
f ( x0 )
( x  x0 ) 
( x  x0 ) 2  ... 
1!
2!
f ( n ) ( x0 )

( x  x0 ) n  Rn ( x),
n!
f ( x)  f ( x0 ) 
f ( n1)  x0   ( x  x0 ) 
 x  x0 n1 , 0    1 
bu erda Rn ( x) 
(n  1)!
Teylor formulasining qoldiq hadi.
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