题目内容
化简:
(1)
-
(2)(
-
)÷
.
(1)
| 12 |
| m2-9 |
| 2 |
| m-3 |
(2)(
| x2-4 |
| x2-4x+4 |
| x-2 |
| x+2 |
| x |
| x-2 |
考点:分式的混合运算
专题:
分析:(1)把分式进行通分,然后相减即可;
(2)首先把括号内的式子化简,把除法转化为乘法,然后利用利用分配律计算乘法,最后通分相减即可.
(2)首先把括号内的式子化简,把除法转化为乘法,然后利用利用分配律计算乘法,最后通分相减即可.
解答:解:(1)原式=
-
=
=
=-
;
(2)原式=【
-
】•
=(
-
)•
=
-
=
=
=
.
| 12 |
| (m+3)(m-3) |
| 2(m+3) |
| (m+3)(m-3) |
=
| 12-2(m+3) |
| (m+3)(m-3) |
=
| 6-2m |
| (m+3)(m-3) |
=-
| 2 |
| m+3 |
(2)原式=【
| (x+2)(x-2) |
| (x-2)2 |
| x-2 |
| x+2 |
| x-2 |
| x |
=(
| x+2 |
| x-2 |
| x-2 |
| x+2 |
| x-2 |
| x |
=
| x+2 |
| x |
| (x-2)2 |
| (x+2)x |
=
| (x+2)2-(x-2)2 |
| x(x+2) |
=
| 8x |
| x(x+2) |
=
| 8 |
| x+2 |
点评:本题主要考查分式的混合运算,通分、因式分解和约分是解答的关键.
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