题目内容
先简化,再求值:
÷(
+1),其中x=
+1.
| x |
| x2-2x+1 |
| x+1 |
| x2-1 |
| 2 |
考点:有理数指数幂的化简求值
专题:计算题
分析:运用多项式化简的出原式=
÷
=
÷
=
,代入求解即可.
| x |
| ((x-1)2 |
| x2+x |
| x2-1 |
| x |
| (x-1)2 |
| x |
| x-1 |
| 1 |
| x-1 |
解答:
解:原式=
÷(
+1)=
÷
=
÷
=
,
∵x=
+1.
∴原式=
=
,
| x |
| x2-2x+1 |
| x+1 |
| x2-1 |
| x |
| ((x-1)2 |
| x2+x |
| x2-1 |
| x |
| (x-1)2 |
| x |
| x-1 |
| 1 |
| x-1 |
∵x=
| 2 |
∴原式=
| 1 | ||
|
| ||
| 2 |
点评:本题考查了代数式的化简求值,属于容易题.
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