题目内容
化简
(1)
÷(a+1)-
;(2)(
-
)×
.
(1)
| 2a+2 |
| a-1 |
| a2-1 |
| a2-2a+1 |
| x2 |
| x-2 |
| 4 |
| x-2 |
| 1 |
| x2+2x |
分析:(1)首先把除法转化成乘法运算,把后边的分式进行约分,然后进行分式的加减运算即可求解;
(2)首先计算括号内的式子,然后计算分式的乘法运算即可.
(2)首先计算括号内的式子,然后计算分式的乘法运算即可.
解答:解:(1)原式=
•
-
=
-
=
=
=-1;
(2)原式=
•
=
•
=
.
| 2(a+1) |
| a-1 |
| 1 |
| a+1 |
| (a+1)(a-1) |
| (a-1)2 |
| 2 |
| a-1 |
| a+1 |
| a-1 |
| 2-(a+1) |
| a-1 |
| 1-a |
| a-1 |
(2)原式=
| x2-4 |
| x-2 |
| 1 |
| x(x+2) |
| (x+2)(x-2) |
| x-2 |
| 1 |
| x(x+2) |
| 1 |
| x |
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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