题目内容
(1)(
-
)÷
;
(2)1-(a-
)÷
.
| x+2 |
| x2-2x |
| x-1 |
| x2-4x+4 |
| 4-x |
| x |
(2)1-(a-
| 1 |
| 1-a |
| a2-a+1 |
| a2-2a+1 |
分析:(1)先把括号内的式子进行通分,再把除法转化成乘法,然后约分即可得出答案;
(2)先把括号里面的式子进行通分,再把除法转化成乘法,然后约分即可得出答案.
(2)先把括号里面的式子进行通分,再把除法转化成乘法,然后约分即可得出答案.
解答:解:(1)(
-
)÷
=(
-
)×
=[
-
]×
=
×
=-
;
(2)1-(a-
)÷
=1-
×
=1+(1-a)
=2-a.
| x+2 |
| x2-2x |
| x-1 |
| x2-4x+4 |
| 4-x |
| x |
=(
| x+2 |
| x(x-2) |
| x-1 |
| (x-2)2 |
| x |
| 4-x |
=[
| (x+2)(x-2) |
| x(x-2)2 |
| x(x-1) |
| x(x-2)2 |
| x |
| 4-x |
=
| x-4 |
| x(x-2)2 |
| x |
| 4-x |
=-
| 1 |
| (x-2)2 |
(2)1-(a-
| 1 |
| 1-a |
| a2-a+1 |
| a2-2a+1 |
=1-
| a(1-a)-1 |
| 1-a |
| (a-1)2 |
| a2-a+1 |
=1+(1-a)
=2-a.
点评:此题考查了分式的混合运算,掌握分式的混合运算法则是本题的关键,用到的知识点是通分、因式分解和约分.
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