题目内容
(1)
•
(2)
÷
(3)
•
(4)
÷
(5)x÷
•x
(6)
÷x•
(7)9a2b÷
•4ab2
(8)
•
÷
(9)
÷(x-y)•
(10)(-
)2•(
)3•(
)4
(11)[
]3•(
)3
(12)(
)2÷
÷(
)2.
| x+y |
| x-y |
| 1 |
| (x+y)2 |
(2)
| a-b |
| a2+ab |
| ab-a2 |
| a2b2-a4 |
(3)
| x2+7x-8 |
| 4x-x3 |
| x2-4 |
| 3x+24 |
(4)
| x2+2xy+y2 |
| xy-y2 |
| xy+y2 |
| x2-2xy+y2 |
(5)x÷
| 1 |
| x |
(6)
| x3+x2 |
| 1-x2 |
| 1 |
| x |
(7)9a2b÷
| 3a |
| 4b |
(8)
| a-4 |
| a2+4a+4 |
| a2+5a+6 |
| a2-a-2 |
| a+3 |
| a-2 |
(9)
| x2+xy |
| x2-xy |
| x2-xy |
| xy |
(10)(-
| a2 |
| b |
| b |
| a |
| 1 |
| ab |
(11)[
| (x-y)2 |
| x+y |
| x2 |
| y2-x2 |
(12)(
| x-1 |
| x2-x-2 |
| x2-2x+1 |
| 2-x |
| 1 |
| x2+x |
分析:(1)直接根据分式的乘法法则进行计算即可;
(2)(4)直接根据分式的除法法则进行计算即可;
(3)根据分式的乘法法则进行计算即可;
(5)、(6)、(7)根据分式的乘法及除法法则进行计算即可;
(8)、(9)、(10)、(11)、(12)根据分式混合运算的法则进行计算即可.
(2)(4)直接根据分式的除法法则进行计算即可;
(3)根据分式的乘法法则进行计算即可;
(5)、(6)、(7)根据分式的乘法及除法法则进行计算即可;
(8)、(9)、(10)、(11)、(12)根据分式混合运算的法则进行计算即可.
解答:解:(1)原式=
=
;
(2)原式=
×
=a-b;
(3)原式=
•
=
;
(4)原式=
×
=
;
(5)原式=x•x•x
=x3;
(6)原式=
×
•
=
;
(7)原式=9a2b•
•4ab2
=48a2b3;
(8)原式=
•
•
=
;
(9)原式=
•
•
=
;
(10)原式=
•
•
=
;
(11)原式=[
•
]3
=[
]3
=-
;
(12)原式=[
]2•
•x2(x+1)2
=
•
•x2(x+1)2
=
.
| 1 |
| (x-y)(x+y) |
| 1 |
| x2-y2 |
(2)原式=
| a-b |
| a(a+b) |
| a2(b+a)(b-a) |
| -a(a-b) |
=a-b;
(3)原式=
| (x-1)(x+8) |
| x(2-x)(2+x) |
| (x+2)(x-2) |
| 3(x+8) |
=
| 1-x |
| 3x |
(4)原式=
| (x+y)2 |
| y(x-y) |
| (x-y)2 |
| y(x+y) |
=
| x2-y2 |
| y2 |
(5)原式=x•x•x
=x3;
(6)原式=
| x2(x+1) |
| (1+x)(1-x) |
| 1 |
| x |
| 1 |
| x |
=
| 1 |
| 1-x |
(7)原式=9a2b•
| 4b |
| 3a |
=48a2b3;
(8)原式=
| a-4 |
| (a+2)2 |
| (a+2)(a+3) |
| (a+1)(a-2) |
| a-2 |
| a+3 |
=
| a-4 |
| (a+2)(a+1) |
(9)原式=
| x(x+y) |
| x(x-y) |
| 1 |
| x-y |
| x(x-y) |
| xy |
=
| x+y |
| y(x-y) |
(10)原式=
| a4 |
| b2 |
| b3 |
| a3 |
| 1 |
| a4b4 |
=
| 1 |
| a3b3 |
(11)原式=[
| (x-y)2 |
| x+y |
| x2 |
| -(x+y)(x-y) |
=[
| x-y |
| -(x+y)2 |
=-
| (x-y)3 |
| (x+y)6 |
(12)原式=[
| x-1 |
| (x+1)(x-2) |
| 2-x |
| (x-1)2 |
=
| (x-1)2 |
| (x+1)2(x-2)2 |
| 2-x |
| (x-1)2 |
=
| x2 |
| 2-x |
点评:本题考查的是分式的乘除法,熟知分式的乘法及除法法则是解答此题的关键.
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