1. Rationalize the denominator of
2
3 6
=
2
3 6
=
2 6
3× 6
=
6
9
×
48
=
=
=
3
48
×
3 6
.
6
6
2. Rationalize the denominator of
3
2
3
.
48
48
48
3× 4 3
48
3
4
1
© 2009 Chung Tai Educational Press. All rights reserved.
3. Rationalize the denominator of
32
=
24
32
24
×
24
24
=
4 2×2 6
24
=
12
3
=
2 3
3
4. Rationalize the denominator of
2 18
108
=
=
=
2 18
108
×
32
.
24
2 18
.
108
108
108
2×3 2 ×6 3
108
6
3
5. Rationalize the denominator of
3 50
.
5 12
3 50 3 50
12
=
×
5 12 5 12
12
=
=
3× 5 2 × 2 3
5 × 12
6
2
2
© 2009 Chung Tai Educational Press. All rights reserved.
6. Rationalize the denominator of
3−4
.
4 3
3−4
3−4
3
=
×
4 3
4 3
3
=
3−4 3
4×3
=
3−4 3
12
7. Rationalize the denominator of
2 2 +7
.
4 2
2 2 +7 2 2 +7
2
=
×
4 2
4 2
2
=
2× 2 + 7 2
4× 2
=
4+7 2
8
8. Rationalize the denominator of
2 7 −3 5
.
35
2 7 −3 5 2 7 −3 5
35
=
×
35
35
35
=
2 245 − 3 175
35
=
2× 7 5 − 3× 5 7
35
=
14 5 − 15 7
35
3
© 2009 Chung Tai Educational Press. All rights reserved.
9. Rationalize the denominator of
3 14 + 8
.
7
3 14 + 8 3 14 + 8
7
=
×
7
7
7
=
3 98 + 56
7
=
3 × 7 2 + 2 14
7
=
21 2 + 2 14
7
10. Rationalize the denominator of
3 6 + 4 24
.
27
3 6 + 4 24 3 6 + 4 × 2 6
=
27
27
=
3 6 +8 6
27
=
11 6
27
×
27
27
=
11 6 × 3 3
27
=
11 × 18
9
=
11 × 3 2
9
=
11 2
3
4
© 2009 Chung Tai Educational Press. All rights reserved.
8
.
2
11. Simplify 9 2 +
9 2+
8
2
8
=9 2+
=9 2+
2
×
2
2
8 2
2
=9 2 +4 2
= 13 2
12. Simplify
6
− 5 20 .
5
6
6
5
− 5 20 =
=
5
8
5
− 5× 2 5
44 5
5
13. Simplify 27 6 −
4 3
5
6 5
− 10 5
5
=−
27 6 −
×
4 3
.
8
= 27 6 −
= 27 6 −
4 3
8
×
8
8
4 3×2 2
8
= 27 6 − 6
= 26 6
5
© 2009 Chung Tai Educational Press. All rights reserved.
14. Simplify 2 24 +
2 24 +
27
27
32
= 2×2 6 +
×
32
32
32
=4 6+
3 3×4 2
32
=4 6+
3 6
8
=
35 6
8
15. Simplify 7 3 p +
7 3p +
12 p
27
12 p
27
, where p > 0.
= 7 3p +
12 p
27
×
27
27
= 7 3p +
12 p × 3 3
27
= 7 3p +
4 3p
3
=
16. Simplify 18a +
18a +
27
.
32
25 3 p
3
5a 7
, where a > 0 .
14a
5a 7
5a 7
14a
= 3 2a +
×
14a
14a
14a
= 3 2a +
5a 98a
14a
5 × 7 2a
14
5 2a
= 3 2a +
2
11 2a
=
2
= 3 2a +
6
© 2009 Chung Tai Educational Press. All rights reserved.
17. Simplify 2 7q −
2 7q −
49 2q
14
49 2q
, where q > 0.
14
= 2 7q −
49 2q
14
×
14
14
49 28q
14
7 × 2 7q
= 2 7q −
2
= 2 7q − 7 7q
= 2 7q −
= − 5 7q
18. Simplify
80b −
80b −
10
, where b > 0.
b 2b
10
10
2b
= 4 5b −
×
b 2b
b 2b
2b
= 4 5b −
20b
b × 2b
= 4 5b −
2 5b
2b 2
=
(4b 2 − 1) 5b
b2
19. Simplify 9 2 + 4 7 ×
9 2+4 7×
63
2 2
63
.
2 2
=9 2+
4 441
2 2
×
2
2
2 × 21 2
2
= 9 2 + 21 2
=9 2+
= 30 2
7
© 2009 Chung Tai Educational Press. All rights reserved.
20. Simplify (2 7 +
(2 7 +
1
2
.
)×
7
3
1
2
1
7
2
)×
= (2 7 +
×
)×
7
3
7
7
3
= (2 7 +
21. Simplify (
(
7
2
)×
7
3
=
15 7
2
×
7
3
=
30 7
7 3
=
30 7
3
×
7 3
3
=
30 21
7×3
=
10 21
7
31 3
6
5
.
−
)×
7
21
27
31 3
6
5
31 3
7
6
21
5
27
−
)×
=(
×
−
×
)×
×
7
21
27
7
7
21
21
27
27
=(
31 21 6 21
5 ×3 3
−
)×
7
21
27
=
29 21
15
×
7
9
=
29 × 315
7×9
=
29 × 3 35
63
=
29 35
21
8
© 2009 Chung Tai Educational Press. All rights reserved.
22. Simplify (3 5 − 7 2 ) ×
(3 5 − 7 2 ) ×
23. Simplify (
(
3
10
=
3
.
10
3(3 5 − 7 2 )
10
×
10
10
=
9 50 − 21 20
10
=
9 × 5 2 − 21× 2 5
10
=
45 2 − 42 5
10
1
4
.
+ 20 − 12 ) ×
3
15
1
4
1
3
4
+ 20 − 12 ) ×
=(
×
+ 2 5 − 2 3) ×
3
15
3
3
15
=(
3
4
+ 2 5 − 2 3) ×
3
15
=
6 5 −5 3
4
×
3
15
=
4(6 5 − 5 3 )
15
×
3 15
15
=
4(6 75 − 5 45 )
45
=
4(6 × 5 3 − 5 × 3 5 )
45
=
8 3−4 5
3
9
© 2009 Chung Tai Educational Press. All rights reserved.
24. Simplify
4 3 − 3 1+ 3
.
−
12
3
4 3 − 3 1+ 3 4 3 − 3
12 1 + 3
3
−
=
×
−
×
12
3
12
12
3
3
4 3 × 2 3 − 3× 2 3
3 +3
−
12
3
=
24 − 6 3 − 4 3 − 12
12
6−5 3
=
6
=
25. Simplify
2 5 −3 2 5 5 −3 2
.
+
20
45
2 5 −3 2
20
+
5 5 −3 2
45
=
2 5 −3 2
20
×
20
20
+
5 5 −3 2
45
×
45
45
=
2 5 ×2 5 −3 2 ×2 5 5 5 ×3 5 −3 2 ×3 5
+
20
45
=
20 − 6 10 75 − 9 10
+
20
45
=
8
10
−
3
2
=
16 − 3 10
6
26. Solve the equation 7 2 x − 12 = 3 2 x , rationalize the denominator and simplify the root obtained.
7 2 x − 12 = 3 2 x
(7 2 − 3 2 ) x = 12
4 2 x = 12
x=
=
=
12
4 2
3
2
×
2
2
3 2
2
10
© 2009 Chung Tai Educational Press. All rights reserved.
27. Solve the equation 2 7 x −
obtained.
2 7x −
2 7x −
1
x = 52 , rationalize the denominator and simplify the root
7
1
x = 52
7
1
7
x×
= 52
7
7
(2 7 −
7
) x = 52
7
13 7
x = 52
7
x = 52 ×
= 4×
7
13 7
7
7
×
7
7
=4 7
28. Solve the equation 3 3 x +
obtained.
3 3x +
3 3x +
6
x = 10 5 , rationalize the denominator and simplify the root
3
6
x = 10 5
3
6
3
x×
= 10 5
3
3
3 3x +
6 3
x = 10 5
3
(3 3 + 2 3 ) x = 10 5
5 3 x = 10 5
x=
10 5
5 3
=
2 5
3
×
3
3
=
2 15
3
11
© 2009 Chung Tai Educational Press. All rights reserved.
6
10
7
x=
x − , rationalize the denominator and simplify the root
2
20
2 2
29. Solve the equation
obtained.
6
10
7
x=
x−
2
20
2 2
6
20
10
2 7
x×
=
x×
−
20
20 2 2
2 2
6 20
20
7
x=
x−
20
2× 2
2
3× 2 5
2 5
7
x−
x=−
10
4
2
(
3 5
5
7
−
)x = −
5
2
2
5
7
x=−
10
2
7 10
x = (− ) ×
2
5
= (−
=−
35
5
)×
5
5
35 5
5
= −7 5
30. Simplify a a3 +
a a3 +
a2
, where a > 0 .
2a
a2
a2
2a
= a2 a +
×
2a
2a
2a
= a2 a +
=
a 2 2a
2a
( 2a + 2 ) a a
2
12
© 2009 Chung Tai Educational Press. All rights reserved.
48c 5 +
31. Simplify
48c 5 +
6 c
3
− 27c , where c > 0 .
6 c
6 c
3
− 27c = 4c 2 3c +
×
− 3 3c
3
3
3
= 4c 2 3c +
6 3c
− 3 3c
3
= (4c 2 + 2 − 3) 3c
= (4c 2 − 1) 3c
32. Rationalize the denominator of
1
=
7 −1
=
1
.
7 −1
1
7 +1
×
7 −1
7 +1
7 +1
( 7 ) 2 − 12
7 +1
6
=
33. Rationalize the denominator of
4
=
5− 3
=
4
5− 3
.
4
5+ 3
×
5− 3
5+ 3
4( 5 + 3 )
( 5 )2 − ( 3 )2
4( 5 + 3 )
2
=2 5+2 3
=
13
© 2009 Chung Tai Educational Press. All rights reserved.
34. Rationalize the denominator of
1
.
3 5−2 3
1
1
3 5+2 3
=
×
3 5−2 3 3 5−2 3 3 5+2 3
=
3 5+2 3
(3 5 ) 2 − (2 3 ) 2
=
3 5+2 3
9×5 − 4×3
=
3 5+2 3
33
35. Rationalize the denominator of
6
=
3+ 2
=
6
3+ 2
.
6
3− 2
×
3+ 2
3− 2
6( 3 − 2)
( 3 )2 − ( 2 )2
= 18 − 12
=3 2 −2 3
36. Rationalize the denominator of
8− 2
.
2− 2
8− 2 2 2− 2
=
2− 2
2− 2
=
2 2 − 2 2+ 2
×
2− 2
2+ 2
=
4 2 + 2× 2 − 2 2 − 2
22 − ( 2 )2
2 2+2
2
= 2 +1
=
14
© 2009 Chung Tai Educational Press. All rights reserved.
37. Rationalize the denominator of
21 − 2 3
.
3 7 +4
21 − 2 3
21 − 2 3 3 7 − 4
=
×
3 7 +4
3 7 +4
3 7 −4
=
3 147 − 4 21 − 6 21 + 8 3
(3 7 ) 2 − 4 2
=
3 × 7 3 − 4 21 − 6 21 + 8 3
63 − 16
=
21 3 − 4 21 − 6 21 + 8 3
47
=
29 3 − 10 21
47
38. Rationalize the denominator of
3 5 +5 3
.
3 5 −5 3
3 5 +5 3 3 5 +5 3 3 5 +5 3
=
×
3 5 −5 3 3 5 −5 3 3 5 +5 3
=
(3 5 ) 2 + 2(3 5 )(5 3 ) + (5 3 ) 2
(3 5 ) 2 − (5 3 ) 2
=
9 × 5 + 30 15 + 25 × 3
9 × 5 − 25 × 3
=
120 + 30 15
−30
= − 4 − 15
15
© 2009 Chung Tai Educational Press. All rights reserved.
39. Rationalize the denominator of
4 2 + 7 12
.
3 2 −2 3
4 2 + 7 12 4 2 + 7 × 2 3
=
3 2 −2 3
3 2 −2 3
40. Simplify
=
4 2 + 14 3 3 2 + 2 3
×
3 2 −2 3 3 2 +2 3
=
12 × 2 + 8 6 + 42 6 + 28 × 3
(3 2 ) 2 − (2 3 ) 2
=
108 + 50 6
18 − 12
=
108 + 50 6
6
=
54 + 25 6
3
2
2
.
−
3 +1
3 −1
2
2
2
3 −1
2
3 +1
−
=
×
−
×
3 +1
3 −1
3 +1
3 −1
3 −1
3 +1
=
2( 3 − 1) − 2( 3 + 1)
( 3 ) 2 − 12
=
2 3−2−2 3−2
3 −1
=
−4
2
= −2
16
© 2009 Chung Tai Educational Press. All rights reserved.
41. Simplify
6
6
.
+
3− 2
3+ 2
6
6
+
=
3− 2
3+ 2
6
3+ 2
6
3− 2
×
+
×
3− 2
3+ 2
3+ 2
3− 2
=
6( 3 + 2) + 6( 3 − 2)
( 3)2 − ( 2 )2
=
18 + 12 + 18 − 12
3− 2
= 2 18
=6 2
42. Simplify
4
5
.
−
4− 5
5+4
4
5
4
4+ 5
5
4− 5
−
=
×
−
×
4− 5
5 +4 4− 5 4+ 5 4+ 5 4− 5
=
4(4 + 5 ) − 5 (4 − 5 )
42 − ( 5 ) 2
16 + 4 5 − 4 5 + 5
16 − 5
21
=
11
=
43. Simplify
3
11 + 3
3
11
.
+
11 + 3
3 − 11
+
11
3 − 11
=
=
3
11 + 3
×
11 − 3
11 − 3
−
11
11 − 3
×
11 + 3
11 + 3
3 ( 11 − 3 ) − 11( 11 + 3 )
( 11) 2 − ( 3 ) 2
33 − 3 − 11 − 33
11 − 3
−14
=
8
7
=−
4
=
17
© 2009 Chung Tai Educational Press. All rights reserved.
44. Simplify
12
3
.
+
8+ 3
8− 3
12
3
2 3
3
+
=
+
8+ 3
8− 3 2 2+ 3 2 2− 3
45. Simplify
=
2 3
2 2− 3
3
2 2+ 3
×
+
×
2 2+ 3 2 2− 3 2 2− 3 2 2+ 3
=
2 3 (2 2 − 3 ) + 3 (2 2 + 3 )
(2 2 ) 2 − ( 3 ) 2
=
4 6 − 2×3+ 2 6 + 3
4×2 −3
=
6 6 −3
5
2 7 −7 2 2 7 +7 2
.
+
2 7 +7 2 2 7 −7 2
2 7 −7 2 2 7 +7 2 2 7 −7 2 2 7 −7 2 2 7 +7 2 2 7 +7 2
+
=
×
+
×
2 7 +7 2 2 7 −7 2 2 7 +7 2 2 7 −7 2 2 7 −7 2 2 7 +7 2
=
[(2 7 ) 2 − 2(2 7 )(7 2 ) + (7 2 ) 2 ] + [(2 7 ) 2 + 2(2 7 )(7 2 ) + (7 2 ) 2 ]
( 2 7 ) 2 − (7 2 ) 2
=
2(2 7 ) 2 + 2(7 2 ) 2
( 2 7 ) 2 − (7 2 ) 2
=
2 × 4 × 7 + 2 × 49 × 2
4 × 7 − 49 × 2
=
252
−70
=−
18
5
18
© 2009 Chung Tai Educational Press. All rights reserved.
46. Simplify
3 2 +5 5−3 2
.
+
3 2 −5 5+3 2
3 2 +5 5−3 2 3 2 +5 3 2 −5
+
=
−
3 2 −5 5+3 2 3 2 −5 3 2 +5
=
3 2 +5 3 2 +5 3 2 −5 3 2 −5
×
−
×
3 2 −5 3 2 +5 3 2 +5 3 2 −5
=
[(3 2 ) 2 + 2(3 2 )(5) + 52 ] − [(3 2 ) 2 − 2(3 2 )(5) + 52 ]
(3 2 ) 2 − 52
=
2(30 2 )
9 × 2 − 25
=−
60 2
7
47. Solve the equation 7 3x = 12 + 3 2 x , rationalize the denominator and simplify the root
obtained.
7 3 x = 12 + 3 2 x
(7 3 − 3 2 ) x = 12
x=
12
7 3 −3 2
=
12
7 3 +3 2
×
7 3 −3 2 7 3 +3 2
=
12(7 3 + 3 2 )
(7 3 ) 2 − (3 2 ) 2
=
84 3 + 36 2
49 × 3 − 9 × 2
=
84 3 + 36 2
129
=
28 3 + 12 2
43
19
© 2009 Chung Tai Educational Press. All rights reserved.
48. Solve the equation 3 5 x + 4 =
obtained.
3 5x + 4 =
1
x
2
3 5x + 4 =
1
2
x×
2
2
3 5x + 4 =
2
x
2
(3 5 −
(
1
x , rationalize the denominator and simplify the root
2
2
) x = −4
2
6 5− 2
) x = −4
2
x = (− 4) ×
2
6 5− 2
=
−8
6 5+ 2
×
6 5− 2 6 5+ 2
=
−8(6 5 + 2 )
(6 5 ) 2 − ( 2 ) 2
=
− 48 5 − 8 2
36 × 5 − 2
=
− 48 5 − 8 2
178
=
−24 5 − 4 2
89
49. Solve the equation 2 10x − 3 6 = 4 15x, rationalize the denominator and simplify the root
obtained.
2 10 x − 3 6 = 4 15 x
(4 15 − 2 10 ) x = −3 6
20
© 2009 Chung Tai Educational Press. All rights reserved.
∴
x=−
3 6
2(2 15 − 10 )
=−
3 6
2 15 + 10
×
2(2 15 − 10 ) 2 15 + 10
=−
6 90 + 3 60
2[(2 15 ) 2 − ( 10 ) 2 ]
=−
6 × 3 10 + 3 × 2 15
2(4 × 15 − 10)
=−
18 10 + 6 15
100
=−
9 10 + 3 15
50
50. Solve the equation ( x + 3) 2 =
obtained.
( x + 3) 2 =
20
3
−
x
5
5
2x + 3 2 =
20
5
3
5
×
−
x×
5
5
5
5
2x + 3 2 =
20 5 3 5
−
x
5
5
( 2+
(
20
3
−
x , rationalize the denominator and simplify the root
5
5
3 5
)x = 4 5 − 3 2
5
5 2 +3 5
)x = 4 5 − 3 2
5
x = (4 5 − 3 2 ) ×
5
5 2 +3 5
=
5(4 5 − 3 2 ) 5 2 − 3 5
×
5 2 +3 5
5 2 −3 5
=
5(20 10 − 12 × 5 − 15 × 2 + 9 10 )
(5 2 ) 2 − (3 5 ) 2
=
5(29 10 − 90)
25 × 2 − 9 × 5
=
5(29 10 − 90)
5
= 29 10 − 90
21
© 2009 Chung Tai Educational Press. All rights reserved.
3
18
1
x+7 3 =
−
x, rationalize the denominator and simplify the root
7
3
3
51. Solve the equation
obtained.
3
18
1
x+7 3 =
−
x
7
3
3
3
7
18
3
1
3
x×
+7 3 =
×
−
x×
7
7
3
3
3
3
3 7
18 3
3
x+7 3 =
−
x
7
3
3
(
3 7
3
+
)x = 6 3 − 7 3
7
3
9 7 +7 3
x=− 3
21
x = (− 3 ) ×
21
9 7 +7 3
=−
21 3
9 7 −7 3
×
9 7 +7 3 9 7 −7 3
=−
189 21 − 147 × 3
(9 7 ) 2 − (7 3 ) 2
=−
189 21 − 441
81× 7 − 49 × 3
=−
189 21 − 441
420
=
21 − 9 21
20
52. If x = 3 − 8 , find the values of the following expressions and simplify the answers obtained.
1
1
(a) x +
(b) x 2 + 2
x
x
(a) x +
1
1
= 3− 8 +
x
3− 8
= 3− 8 +
1
3+ 8
×
3− 8 3+ 8
= 3− 8 +
3+ 8
3 − ( 8 )2
2
= 3− 8 +3+ 8
=6
22
© 2009 Chung Tai Educational Press. All rights reserved.
1
1
1
(b) ( x + ) 2 = x 2 + 2( x)( ) + 2
x
x
x
= x2 + 2 +
x2 +
∴
1
x2
1
1
= ( x + )2 − 2
2
x
x
= 62 − 2
= 34
53. If y = 6 + 5 , find the values of the following expressions and simplify the answers
obtained.
(a)
1
y
y−
(a) y −
(b)
y2 +
1
y2
(c)
y2 −
1
y2
1
1
= 6+ 5−
y
6+ 5
= 6+ 5−
= 6+ 5−
1
6− 5
×
6+ 5
6− 5
6− 5
( 6 )2 − ( 5 )2
= 6+ 5− 6+ 5
=2 5
(b) ( y −
1 2
1
1
) = y 2 − 2( y )( ) + 2
y
y
y
= y2 − 2 +
∴
y2 +
1
y2
1
1
= ( y − )2 + 2
2
y
y
= (2 5 ) 2 + 2
= 22
23
© 2009 Chung Tai Educational Press. All rights reserved.
2
(c) y −
1
1
1
= ( y − )( y + )
2
y
y
y
1
1
= ( y − )( 6 + 5 +
)
y
6+ 5
= (2 5 )( 6 + 5 +
= (2 5 )[ 6 + 5 +
1
6− 5 [
×
) From the result of (a) ]
6+ 5
6− 5
6− 5
]
( 6 )2 − ( 5 )2
= (2 5 )( 6 + 5 + 6 − 5 )
= (2 5 )(2 6 )
= 4 30
54. If α = 2 + 3 and β = 2 − 3 , find the values of the following expressions and simplify the
answers obtained.
(a)
1
α
(a)
1
1
=
α 2+ 3
(b)
=
1
2− 3
×
2+ 3 2− 3
=
2− 3
2 − ( 3 )2
1
β
(c)
1
1
− 2
2
α
β
2
=2− 3
(b) ∵
∴
(c)
1
= β [ From the result of (a) ]
α
1 1
= = α = 2+ 3
β α1
1
1
1 1 1 1
− 2 = ( + )( − )
2
α β α β
α
β
= [(2 − 3 ) + (2 + 3 )][(2 − 3 ) − (2 + 3 )] [From the results of (a) and (b)]
= (4)(−2 3 )
= −8 3
24
© 2009 Chung Tai Educational Press. All rights reserved.
55. If α = 1 + 5 and β = 3 − 5 , find the values of the following expressions and simplify the
answers obtained.
(a)
1 1
+
α β
(a)
1 1
1
1
+ =
+
α β 1+ 5 3 − 5
(b)
1 1
−
α β
=
1
1− 5
1
3+ 5
×
+
×
1+ 5 1− 5 3 − 5 3 + 5
=
1− 5
3+ 5
+ 2
2
1 − ( 5)
3 − ( 5 )2
(c)
1
1
− 2
2
α
β
2
− (1 − 5 ) + 3 + 5
4
2+2 5
=
4
1+ 5
=
2
=
(b)
1 1
1
1
− =
−
α β 1+ 5 3 − 5
=
1
1− 5
1
3+ 5
×
−
×
1+ 5 1− 5 3 − 5 3 + 5
=
1− 5
3+ 5
− 2
2
1 − ( 5)
3 − ( 5 )2
2
− (1 − 5 ) − (3 + 5 )
4
−4
=
4
= −1
=
(c)
1
1
1 1 1 1
− 2 = ( + )( − )
2
α β α β
α
β
1+ 5
)(−1) [ From the results of (a) and (b) ]
2
1+ 5
=−
2
=(
25
© 2009 Chung Tai Educational Press. All rights reserved.
1
1
and q =
, find the values of the following expressions and simplify the
2− 3
2+ 3
answers obtained.
56. If p =
(a)
p+q
(a)
p+q =
(b)
=
p−q
1
1
+
2− 3 2+ 3
1
2+ 3
1
2− 3
×
+
×
2− 3 2+ 3 2+ 3 2− 3
2+ 3 +2− 3
22 − ( 3 ) 2
=4
=
(b)
p−q=
1
1
−
2− 3 2+ 3
=
1
2+ 3
1
2− 3
×
−
×
2− 3 2+ 3 2+ 3 2− 3
=
2+ 3−2+ 3
22 − ( 3 )2
=2 3
(c)
p 2 − q 2 = ( p + q)( p − q )
= 4(2 3 ) [ From the results of (a) and (b) ]
=8 3
57. Simplify
2 x −1
, where x > 1.
x +1 − x −1
2 x −1
=
x +1 − x −1
= 2[
2 x −1
x +1 + x −1
×
x +1 − x −1
x +1 + x −1
( x − 1)( x + 1) + ( x − 1)
]
( x + 1) − ( x − 1)
= x2 − 1 + x − 1
26
© 2009 Chung Tai Educational Press. All rights reserved.
(c)
p2 − q2
58. Simplify
4 y−4 −4 y+4
y+4 + y−4
4 y−4 −4 y+4
y+4+
y−4
, where y > 4.
y+4 − y−4
= ( − 4) ×
y+4+
y−4
y+4 − y−4
×
y+4 − y−4
= ( − 4) ×
( y + 4 − y − 4 )2
( y + 4) − ( y − 4)
= ( − 4) ×
( y + 4) − 2 ( y + 4)( y − 4) + ( y − 4)
8
=−
2 y − 2 y 2 − 16
2
=
y 2 − 16 − y
59. Simplify the following, where m > 0 and n > 0 .
(a)
(a)
m
4
−
m −4
m+4
m
m −4
−
4
m +4
(b)
=
=
(b)
7
2 m −n 3
+
m
=
m −4
×
m +4
m +4
−
7
5
+
2 m −n 3 2 m +n 3
4
m +4
×
m −4
m −4
m + 4 m − (4 m − 16)
m − 42
m + 16
m − 16
5
2 m +n 3
=
=
=
7
2 m −n 3
×
2 m +n 3
2 m +n 3
+
5
2 m +n 3
×
2 m −n 3
2 m −n 3
14 m + 7n 3 + 10 m − 5n 3
(2 m ) 2 − (n 3 ) 2
24 m + 2n 3
4m − 3n 2
27
© 2009 Chung Tai Educational Press. All rights reserved.
60. Simplify the following, where m > n > 0.
n3
n3
(a)
3m − 2n − 3m + 2n
3m − 2n + 3m + 2n
(a)
3m − 2n − 3m + 2n
=
3m − 2n + 3m + 2n
3m − 2n − 3m + 2n
3m − 2n − 3m + 2n
×
3m − 2n + 3m + 2n
3m − 2n − 3m + 2n
=
(3m − 2n) − 2 (3m − 2n)(3m + 2n) + (3m + 2n)
(3m − 2n) − (3m + 2n)
=
6m − 2 9m 2 − 4n 2
− 4n
=
9m 2 − 4n 2 − 3m
2n
(b)
n3
m − m2 − n2
=
=
+
n3
m − m2 − n2
(b)
m − m2 − n2
+
m + m2 − n2
n3
m + m2 − n2
×
m + m2 − n2
m + m2 − n2
n3
+
m + m2 − n2
×
m − m2 − n2
m − m2 − n2
mn3 + n3 m 2 − n 2 + mn3 − n3 m 2 − n 2
m 2 − (m 2 − n 2 )
2mn3
n2
= 2mn
=
61. Prove that
x +1
x
−
≡ 2 x + 1, where x > 0 .
x +1 − x
x +1 + x
L.H.S. =
x +1
−
x +1 − x
x
x +1 + x
=
x +1
×
x +1 − x
x +1 + x
−
x +1 + x
x
×
x +1 + x
[ x + 1 + x( x + 1) ] − [ x( x + 1) − x]
( x + 1) − x
= 2x + 1
=
= R.H.S.
∴
x +1
x
−
≡ 2x + 1
x +1 − x
x +1 + x
28
© 2009 Chung Tai Educational Press. All rights reserved.
x +1 − x
x +1 − x
62. Prove that
x−3
+
x+3 − x−3
x−3
+
x+3 − x−3
L.H.S. =
x+3
x
≡ , where x > 3.
x+3 + x−3 3
x+3
x+3 + x−3
x−3
x+3 + x−3
×
+
x+3 − x−3
x+3 + x−3
=
x+3
x+3 − x−3
×
x+3 + x−3
x+3 − x−3
( x − 3)( x + 3) + ( x − 3) + ( x + 3) − ( x + 3)( x − 3)
( x + 3) − ( x − 3)
2x
=
6
x
=
3
= R.H.S.
=
∴
x−3
+
x+3 − x−3
63. Prove that
L.H.S. =
=
x+3
x
≡
x+3 − x−3 3
3x + 2
x−4
2x − 1
, where x > 4 .
+
≡
x+3
3x + 2 + x − 4
3x + 2 − x − 4
3x + 2
x−4
+
3x + 2 + x − 4
3x + 2 − x − 4
3x + 2
3x + 2 − x − 4
x−4
3x + 2 + x − 4
×
+
×
3x + 2 + x − 4
3x + 2 − x − 4
3x + 2 − x − 4
3x + 2 + x − 4
(3 x + 2) − (3 x + 2)( x − 4) + ( x − 4)(3 x + 2) + ( x − 4)
(3 x + 2) − ( x − 4)
4x − 2
=
2x + 6
=
=
2x − 1
x+3
= R.H.S.
∴
3x + 2
x−4
2x − 1
+
≡
x+3
3x + 2 + x − 4
3x + 2 − x − 4
29
© 2009 Chung Tai Educational Press. All rights reserved.
64. (a) Simplify
1
2k + 2k + 4
, where k ≥ 0 .
(b) Hence find the values of the following.
1
1
1
1
+
+
+!+
2+ 6
6 + 10
10 + 14
30 + 34
1
1
1
1
+
+
+!+
2+ 6
4+ 8
6 + 10
32 + 36
(i)
(ii)
(a)
1
2k + 2k + 4
1
2k + 2k + 4
=
2k − 2k + 4
2k − (2k + 4)
=
2k − 2k + 4
−4
=
2k + 4 − 2k
4
×
2k − 2k + 4
2k − 2k + 4
1
1
1
1
+
+
+!+
2+ 6
6 + 10
10 + 14
30 + 34
(b) (i)
(ii)
=
=
6− 2
10 − 6
14 − 10
34 − 30
+
+
+!+
4
4
4
4
=
34 − 2
4
[ From the result of (a) ]
1
1
1
1
+
+
+!+
2+ 6
4+ 8
6 + 10
32 + 36
1
1
1
1
=(
+
+
+!+
)
2+ 6
6 + 10
10 + 14
30 + 34
1
1
1
1
+(
+
+
+!+
)
4+ 8
8 + 12
12 + 16
32 + 36
=
34 − 2
36 − 4
+
4
4
=
34 − 2 + 4
4
30
© 2009 Chung Tai Educational Press. All rights reserved.
65. (a) Simplify
1
, where k ≥ 2 .
3k − 6 + 3k
(b) Hence find the values of the following.
1
1
1
1
+
+
+!+
9 + 15
15 + 21
21 + 27
75 + 81
1
1
1
1
1
−
+
−
+!+
6 + 12
9 + 15
12 + 18
15 + 21
78 + 84
(i)
(ii)
(a)
1
=
3k − 6 + 3k
1
3k − 6 − 3k
×
3k − 6 + 3k
3k − 6 − 3k
=
3k − 6 − 3k
(3k − 6) − 3k
=
3k − 6 − 3k
−6
=
3k − 3k − 6
6
1
1
+
+
9 + 15
15 + 21
(b) (i)
=
15 − 9
+
6
=
81 − 9
6
21 − 15
+
6
1
1
+!+
21 + 27
75 + 81
27 − 21
81 − 75
[ From the result of (a) ]
+!+
6
6
=1
(ii)
1
1
1
1
1
−
+
−
+!+
6 + 12
9 + 15
12 + 18
15 + 21
78 + 84
1
1
1
1
=(
+
+
+!+
)
6 + 12
12 + 18
18 + 24
78 + 84
1
1
1
1
−(
+
+
+!+
)
9 + 15
15 + 21
21 + 27
75 + 81
=(
12 − 6
18 − 12
24 − 18
84 − 78
+
+
+!+
) −1
6
6
6
6
=
84 − 6
−1
6
=
2 21 − 6 − 6
6
[ From the results of (a) and (b)(i) ]
31
© 2009 Chung Tai Educational Press. All rights reserved.
x2 − 1
66. Prove that
2
2x − 2 + x − 1
L.H.S. =
=
=
x2 − 1
2x − 2 + x2 − 1
x2 − 1
2x − 2 + x2 − 1
−
×
−
2x − 2
2
2x − 2 − x − 1
≡
x+3
, where x > 1.
x −1
2x − 2
2x − 2 − x2 − 1
2x − 2 − x2 − 1
2x − 2 − x2 − 1
−
2x − 2
2x − 2 − x2 − 1
×
2x − 2 + x2 − 1
2x − 2 + x2 − 1
( x 2 − 1)(2 x − 2) − ( x 2 − 1) − (2 x − 2) − (2 x − 2)( x 2 − 1)
(2 x − 2) − ( x 2 − 1)
=
− x2 − 2x + 3
− x2 + 2x − 1
=
x2 + 2x − 3
x2 − 2x + 1
=
( x + 3)( x − 1)
( x − 1) 2
=
x+3
x −1
= R.H.S.
∴
x2 − 1
2x − 2 + x2 − 1
67. (a) Simplify
−
2x − 2
2x − 2 − x2 − 1
≡
x+3
x −1
2b
, where a > b > 0 .
a−b + a+b
(b) Hence find the value of the following.
1
1
1
1
+
+
+!+
4 + 14
14 + 24
24 + 34
134 + 144
(c) Given that
c
c
c
c
+
+
+!+
= 5, find the value of c.
4+ 8
8 + 12
12 + 16
140 + 144
32
© 2009 Chung Tai Educational Press. All rights reserved.
2b
=
a−b + a+b
(a)
2b
a−b − a+b
×
a−b + a+b
a−b − a+b
=
2b( a − b − a + b )
( a − b) − ( a + b)
=
2b( a − b − a + b )
−2b
= a+b − a−b
10
10
10
10
+
+
+!+
4 + 14
14 + 24
24 + 34
134 + 144
(b)
= ( 14 − 4 ) + ( 24 − 14 ) + ( 34 − 24 ) + ! + ( 144 − 134 )
[ From the result of (a) ]
= 144 − 4
= 10
∴
1
1
1
1
+
+
+!+
=1
4 + 14
14 + 24
24 + 34
134 + 144
c
c
c
c
+
+
+!+
=5
4+ 8
8 + 12
12 + 16
140 + 144
(c)
c
2(2)
2(2)
2(2)
2(2)
[
+
+
+!+
]=5
4 4+ 8
8 + 12
12 + 16
140 + 144
c
[( 8 − 4 ) + ( 12 − 8 ) + ( 16 − 12 ) + ! + ( 144 − 140 )] = 5
4
[ From the result of
c( 144 − 4 ) = 20
10c = 20
c=2
(a) ]
68. If α = 7 + 3 5 and β = 7 − 3 5 , find the values of the following expressions and simplify the
answers obtained.
33
© 2009 Chung Tai Educational Press. All rights reserved.
(a)
α
β
(a)
α 7+3 5
=
β 7−3 5
(b)
(b)
=
7+3 5 7+3 5
×
7−3 5 7+3 5
=
7 2 + 2(7)(3 5 ) + (3 5 ) 2
7 2 − (3 5 ) 2
=
49 + 42 5 + 9 × 5
49 − 9 × 5
=
94 + 42 5
4
=
47 + 21 5
2
β
α
β
2
[ From the result of (a) ]
=
α 47 + 21 5
2
47 − 21 5
=
×
47 + 21 5 47 − 21 5
=
2(47 − 21 5 )
472 − (21 5 ) 2
=
94 − 42 5
2 209 − 441 × 5
=
94 − 42 5
4
=
47 − 21 5
2
34
© 2009 Chung Tai Educational Press. All rights reserved.
(c)
α 2 + β2
αβ
(c)
α 2 + β2 α 2 β2
=
+
αβ
αβ αβ
=
α β
+
β α
=
47 + 21 5 47 − 21 5
+
2
2
= 47
1
and β = 3 − 7 , find the values of the following expressions and simplify the
3+ 7
answers obtained.
(a) αβ
(b) α + β
(c) α 2 + β2
69. If α =
(a) αβ =
1
× (3 − 7 )
3+ 7
=
3− 7 3− 7
×
3+ 7 3− 7
=
32 − 2(3)( 7 ) + ( 7 ) 2
32 − ( 7 ) 2
9−6 7 +7
9−7
16 − 6 7
=
2
=8−3 7
=
(b) α + β =
1
+ (3 − 7 )
3+ 7
=
1
3− 7
×
+3− 7
3+ 7 3− 7
=
3− 7
+3− 7
3 − ( 7 )2
2
3− 7
+3− 7
2
9−3 7
=
2
=
(c) ∵
(α + β) 2 = α 2 + 2αβ + β2
35
© 2009 Chung Tai Educational Press. All rights reserved.
α 2 + β 2 = (α + β) 2 − 2αβ
∴
9−3 7 2
) − 2(8 − 3 7 )
2
9 2 − 2(9)(3 7 ) + (3 7 ) 2
− 16 + 6 7
4
81 − 54 7 + 63 − 64 + 24 7
4
80 − 30 7
4
40 − 15 7
2
=(
=
=
=
=
1
1
, where a > 1, find the values of the following expressions and simplify
=
a 2− 3
the answers obtained.
70. If
a+
(a) a +
1
a
a+
(a)
(b)
a−
1
1
=
a 2− 3
=
1
2+ 3
×
2− 3 2+ 3
=
2+ 3
4−3
=2+ 3
1 2
1
1 2
) = ( a ) 2 + 2( a )( ) + (
)
a
a
a
1
=a+2+
a
1
1 2
a+ =( a +
) −2
a
a
( a+
∴
= (2 + 3 ) 2 − 2
= 2 2 + 2(2)( 3 ) + ( 3 ) 2 − 2
= 4+ 4 3 +3−2
=5+4 3
36
© 2009 Chung Tai Educational Press. All rights reserved.
1
a
(b) ( a −
1 2
1
1
) = ( a ) 2 − 2( a )( ) + ( ) 2
a
a
a
1
=a+ −2
a
=5+4 3 −2
= 3+ 4 3
∵
a >1
a >1
1
<1
a
∴
a−
1
>0
a
∴
a−
1
= 3+ 4 3
a
37
© 2009 Chung Tai Educational Press. All rights reserved.