分数指数幂
1.下列命题中,正确命题的个数是
.
① n n = a 2 0
= 1 a ② 若 a ∈R ,则 (a -a + 1) ③ 3 x + y = x + y ④ 3 - 5=
6
- 5
2
4
3
4
3
2.下列根式、分数指数幂的互化中,正确的序号是
.
1
(x ≠ 0) ②
x x = x 3
③ x - 1 =- 3
3
4
1
x )- 3 = ① - x = (- x)
4 3 x
④ x · x = x
12 ⑤ ( 4
2
y 4
y 3
⑥
6
2
1
(xy ≠ 0)
y =y (y<0)
x
3
c b
3.若 a = 2, b = 3, c =- 2,则 (a ) = __________. 4.根式 a
a 的分数指数幂形式为
.
4 2
5.
- 25 = __________.
- (2k +1)-(2k - 1)-2k
6. 2- 2 + 2 的化简结果是
. 7. (1)设 α, β是方程 2x 2
1 + +3x + 1= 0 的两个根,则 ( ) α β= __________.
4 x y 1 (2)若 10 = 3,10 = 4,则 10x - 2y = __________. 8. (1)求下列各式的值: 2 1 1
4 3 ① 27 ; ②(6 ) ; ③ ( )- .
3 4 2
9 2 -3
1 1 (2)解方程: ① x
=8;② x = 94.
9.求下列各式的值:
2
125 1 7 0.5
(1)(0.027) 3+ ( 27 )3- (29) ;
1 117 13
- 1 3 3 1 -1 (2)(3)2+3·( 3-2) - (164)4- (3 )4- (3) .
11-1
10.已知 a2+a-2= 4,求 a+ a的值.
11.化简下列各式:
2 1
5x-3y2
(1)
1 -1 1 5 1 1
;
-4x y2-6 x3y-6
m+ m -1+ 2
(2)1 1 . m-2
+m 2
.
21
12. [(- 2) ] -2的值是.
36
3
69494
的结果是.13.化简 ( a ) ·( a )
14.以下各式,化简正确的个数是.211
①a5a-3 a-15= 1
6- 92-46
②(a b)-= a b
3
111212
③(- x4y-3)(x-2 y3)(- x4y3)
= y 1 1 3
-15a 2b3c-43
④
1 1 5=-5ac
25a - b c
2 3 4
15. (2010 山东德州模拟, 4 改编 )如果 a3= 3, a10=384 ,则 a3[(a101 n
.a) ] 等于
7
3
16.化简3
a- b
3
a- 2b
2
.+的结果是
17.下列结论中,正确的序号是.
233
①当 a<0 时, (a ) = a
2
②n
a n= |a|(n>1 且 n∈ N * )
10
③函数 y= (x-2)- (3x- 7) 的定义域是 (2,+∞ )
2
④若 100a= 5,10b= 2,则 2a+ b =1
18. (1) 若 a= (2+
- 1-1- 2
+ (b+ 1)
- 2
.3) , b = (2-3),则 (a+ 1)的值是
.
(2)若 x> 0, y> 0,且 x(x+y)= 3y( x+ 5y),则2x+ 2xy+ 3y
的值是.x- xy+ y
11
2 009 n- 2 009 -n*2
+1 +a)n.
19.已知 a=(n∈ N ),则 ( a的值是
2
11111
20.若 S= (1+2-32 )(1+ 2-16)(1+ 2-8)(1+ 2-4)(1+ 2-2),那么 S 等于.21.先化简,再求值:
2535
a · a
(1),其中 a=8 -3;
107
a · a
3x- 3x
a + a2x
(2) a+a
,其中 a = 5.
x- x
22.(易错题 )计算:
3 0- 2 1 10.5
(1)(25) + 2 ·(24)-2- (0.01);
7 0.5- 210 2037 (2)(29) + 0.1+ (227)-3- 3π+48;
17 0-1
[81- 0.253111
(3)(0.008 1) --[3× ( ) ]×+ (3 )- ]-- 10×0.027 .
488323
33
11x2+ x-2+ 2
23.已知 x2+x-2= 3,求x2+x-2+3的值.
24.化简下列各式:
x
- 2
- 2
- 2
- 2
+ y
x
- y
(1)
2
2-
2
2
;
x - 3+ y - 3 x - 3- y - 3
4
1
(2)
a 3-8a 3b
3 b
3 a.
÷(1- 2
)× 2
3 2
a
a + 2 a
b + 4b 3
3
答案与解析
基础巩固
n
n
a ,当 n 为奇数时, 1. 1 ∵ a =
|a|,当 n 为偶数时,
∴① 不正确;
2
1 2
3
∵a ∈ R ,且 a - a + 1= (a - ) + ≠0 ,∴② 正确;
4 3
∵ x + y 为多项式, ∴③ 不正确; ④中左边为负,右边为正显然不正确. ∴只有 ② 正确.
1
2.②⑤ ① - x =- x 2, ∴① 错;
1 1 1 3 1 3
② x x = (x x) = (x ·x ) = (x ) = x , ∴② 对;
2 2 2 2 2 4
1 1 1 ③ x -3= 1=, ∴③ 错;
x 3 3
x
④ 3
4 1 1 1 1 7
x · x = x ·x 4
= x + = x ,
3 3
4
12
∴④ 错;
x
3 y 3
= 4
y 3
⑤( )- = ( )
x ,
y
4 x 4
∴⑤ 对;
⑥ 6
2
1
1
y = |y|3 =- y 3(y<0) , ∴⑥ 错.
∴②⑤ 正确.
3. 1
c b
bc
3×(- 2)
- 6
1 1
(a ) = a
=2 = 2 = 6=.
64
2 64
3 1
1 3
4. a 2 a a = a ·a 2= a1+2= a 2.
5. 5 - 25 = 4 25 = 4
5 = 5.
4 2 2 4
6.- 2
- (2k + 1)
- (2k + 1)
- (2k - 1)
-2k -2k -1
- 2k1
-2k
1 - 2k
1 - 2k
∵ 2
- 2
+2
= 2
·2 - 2 ·2 + 2 =( - 2 + 1)·2 =-
2 ·2
2
=- 2 -(2k + 1).
3
3
7. (1)8
(2)2 (1)由根与系数的关系,得 α+ β=- 2 ,
1 +
1 3 - 2
3 3
∴( ) α β
)- = 2 =8.
= ( )- = (2
4
4 2
2
x
y
1
x
1 x
y 11 3
(2)∵ 10 = 3,10 = 4, ∴ 10x - 2y = 10 ÷102y =10 ÷(10 )2= 3÷42= 2.
2 3 2 2 2
8.解: (1)① 273= (3 )3= 33×3 = 3 = 9.
1 1 25 1
② (64 )2 =( 4 )2
5 2 1
5 1
5 = [( 2) ]2 = (2)2× 2= 2.
4
3
2 3
③ (9)- 2= (3)2× (- 2)
2 - 3
3 3
27 =(3) = (2) = 8 . - 3 1 - 3
(2)①∵ x = 8= 2 , ∴x = 2.
②∵ x = 9 1 , 4
∴( 2 1 2
1 x) = (9 ) = 9 .
4
2 2 1
∴ x =(3 )2= 3.
9.解:
3
2 125 1
25 1 9
5 5 9
(1)原式= (0.3 ) + (
27 ) - (
9 ) =
+ - =.
3
3
2
100
3
3
100
1 3
81 1
2 3
1
(2)原式= 3-2 + 3- 2 - (64)4-(3- 3)4- 3
3
3 4 1 1
= 3 +
3( 3+ 2)- [4(4) ]4 -3 -2- 3
3
3 3
=
3 + 3+ 6- 2 ·- - 3
4 3
6 3
2.
= -4
1 1
10.解: ∵a 2+ a - 2= 4.
∴两边平方,得 a + a -
1+ 2= 16.
∴a + a -
1= 14.
11.解: (1)原式=
24 2 1 1 1 1 0
1 1 × 5× x -
+ 1- × y - + = 24x
y = 24y ;
5
3
3
2
2 6
6
6
(2)原式
1 2
1
1
1 2
m 2 + 2m 2·m - 2+ m - 2
=
1
1
m - 2+ m 2 1
1 2
m 2+ m - 2
1
1
=
1 1 = m 2+ m - 2. m 2+m - 2
能力提升
2
1 1 2
12. 2
原式= 2- 2= 2 = 2 .
4
3
9 4 6
9 4
3 1 4
1 4 1 4 1 4
2 2 4
原式= ( 13. a
a ) ·(
a
) =(a ×
) ·(a3× 6 ) = (a ) ·(a ) =a ·a = a .
6
3
2 3
2
2
14. 3 由分数指数幂的运算法则知 ①②③ 正确;
对④ , ∵ 左边=-
3 1 1 1 1
3 5
3 1 0 - 2 3
- 2
5 a + b
- c - - =-
a b c =- ac ≠ 右边, ∴④ 错误.
2 2
3 3
4
4 5
5
n
384 1 n 1 n
1 n
n
15. 3·2
原式= 3·[( 3 )7] = 3·[(128) 7] =3 ·(27× 7) = 3·2 .
16. b 或 2a - 3b
a -
b + 2b - a , a < 2b
b , a <2b ,
原式= a - b + |a - 2b| =
= 2a - 3b , a ≥ 2b.
a -
b + a - 2b , a ≥ 2b
2
3
2
1 3
3
3 3
17. ④ ①中,当 a < 0 时, (a )2 =[(a )2] =(|a|) = (- a) =- a ,
∴① 不正确;
当 a < 0, n 为奇数时, n
n
a = a ,
∴② 不正确;
x - 2≥ 0, ③中,有
3x - 7≠ 0,
7
即 x ≥ 2 且 x ≠ 3,
7 7
故定义域为 [2, 3)∪ (3 ,+ ∞ ),
∴③ 不正确;
④中, ∵ 100a = 5,10b =2 ,
∴ 102a =5,10 b = 2,102a × 10b = 10.
∴ 2a + b =1.∴④ 正确.
2
1
1
18. (1) 3 (2)3
(1)a = 2 + 3 =2 - 3, b = 2- 3 = 2+ 3 ,
∴(a + 1) -
2 + (b + 1) -
2 = (
3 - 3 ) -
2 + (
3 + 3 ) -
2
=
1 2 + 1 2 =
3 - 3 3+ 3
3 + 3 2+ 3- 3
2
3- 3 2
2
·3+ 3
2
2
3 + 2·3 · 3+ 3+ 3 - 2·3· 3+ 3
= [ 3 - 3 3+ 2
3 ]
2 × 9+ 6 24 2 =
9- 3 2
=36 = 3.
(2)由已知条件,可得
( x)2- 2 xy -15(
y)2= 0,
∴ x + 3 y = 0 或 x -5 y = 0.
∵ x >0, y > 0,
∴ x = 5 y , x =25y.
50y + 2 25y 2+ 3y
∴原式=
2+ y
25y - 25y 50y + 10y + 3y 63y
= = = 3.
25y - 5y + y 21y
1 1
2 009 n - 2 009- n
19. 2 009 ∵ a =
2
,
2
2
∴ a 2
+ 1= 1+
2 009
n +
2 009 - n -
2
4
1 2
1 2
2 009n +2+ 2 009 - n
=
4
11 2 009n+ 2 009 -n2
=() .
2
∴
2
a + 1+ a
1111 2 009 n+ 2 009-n 2 009 n- 2 009 -n
=
2+
2
1
=2 009 n .
2n 1 n
∴( a
+ 1+ a) = (2 009n) = 2 009.
11 -1
20.2(1- 2-32)
原式=
111111 1- 2-32 1+ 2-32 1+ 2 -16 1+ 2-81+ 2-41+ 2-2
1
1 - 2-32
11111
1- 2-16 1+ 2-16 1+ 2-8 1+ 2-4 1+ 2-2
=1
1- 2-32
1111
1- 2-81+ 2-8 1 +2 -4 1 +2 -2
=1
1- 2-32
111
1- 2-41+ 2-4 1 +2 -2
=1
1 -
2 -32
11
1- 2-21+ 2-2
=1
1- 2-32
-1
1 - 21 1 -1
=1=2(1- 2-32) .
1- 2-32
3 71
21.解: (1)原式= a2 +5-10-2
7 5 7
=a5=(8-3)5
737
- 7
1
=8 -3= (2 )-3= 2=
128
.
x 3
-x 3
a + a
(2)原式= x - x
a + a
x - x
2x x -x
- 2x
a + a
a - a ·a
+a
=
x
- x
a + a
2x
-2x
1 1
=a - 1+ a = 5- 1 + = 4 .
5 5
1 4 1 -( 1 1 1
2 1 1 1 1 1
22.解: (1)原式= 1 + ·( ) 100 ) = 1+ × - ( )2× = 1+ - 10 = 1 .
4 9 2 2 4 3 10 2 6 1
5 25 1 1 - 2 64 2 37
(2)原式= ( 9 )2+ (10) +(27)- 3- 3× 1+ 48
5 4 - 2 37
= 3 + 100+ (3 ) - 3+ 48
5
9 37 = 3 + 100+ 16- 3+48= 100.
(3)原式= [(0.3)
41
- 1 4
1 27 1 1 3
1
]- - 3 × [(3 )- + (
8
)- ]- - 10× [(0.3) ]
4
4
3
2
3
- 1
1 - 1
3 -1
1
=0.3 - 3[3 +(2) ]- 2- 10× 0.3
10 1 1 2 1 10 1
= 3 - 3(3+3 )-2 -3 = 3 - 3- 3= 0.
1
1
23.解: ∵x 2 +x - 2= 3, ∴ (x 1
+ x - 1)2
= 9.
22 ∴ x +x -
1= 7.
1 3
1 3
∴原式= x 2 + x -2 + 2
2
- 2
x + x + 3
1 1 -1 + x - x - 1+ x + 2
x 2 2 =
- 1 2
x + x - 2+ 3 3 × 7- 1 + 2 2 =
72- 2+ 3 = 5
.
拓展探究
2 3
2 3
2 3 2 3
x - 3
+ y - 3
x - 3 - y - 3
2 22 2 2 2
24.解: (1)原式=
2 2 -
2
2
=(x - 3) - x -3 ·y - 3+ (y - 3) - (x -
x -3 + y - 3 x -3 -y - 3
2 2
2 2
2 2
2 3) - x - 3·y -3
- (y - 3) =- 2(xy)-3 .
1
1 3 1
3
1
a 3[ a 3 - 2
b 3 ]
b 3
1
(2)原式= 2
1 1
1 2÷(1-
2 1 )× a 3
a 3 +2a 3
b 3+ 2b 3 a 3
1 1 1
2 1 1
a 3 a 3 -2
b 3 [a 3+ 2a 3b 3+
= 2 1 1
1 2
a 3+ 2a 3
b 3+ 2b 3
1 1 1
a 3·a 3 ·a 3= a.
1 2 1 1 1 1 1 1
2b 3 ] a 3- 2b 3 1 a 3 a 3- 2b 3 ·1
a 3
1
÷ 1 ×a = 1 × 1
× a =
3
1
3
a 3
a 3- 2
b 3