题目内容
已知:| x2 |
| x2-2 |
| 1 |
| 1-a |
| 1 |
| 1-x |
| 1 |
| 1+x |
| x |
| x2-1 |
分析:先将分式化简,再由
=
,可得
=a,整体代入即可.
| x2 |
| x2-2 |
| 1 |
| 1-a |
| 2 |
| x2 |
解答:解:(
-
)÷(
+x)
=
÷
=-
•
=-
,
∵
=
,
∴
=1-
=1-a,
∴
=a,
∴原式=a.
| 1 |
| 1-x |
| 1 |
| 1+x |
| x |
| x2-1 |
=
| 1+x-1+x |
| (1-x)(1+x) |
| x(1+x2-1) |
| x2-1 |
=-
| 2x |
| (x+1)(x-1) |
| (x+1)(x-1) |
| x3 |
=-
| 2 |
| x2 |
∵
| x2 |
| x2-2 |
| 1 |
| 1-a |
∴
| x2-2 |
| x2 |
| 2 |
| x2 |
∴
| 2 |
| x2 |
∴原式=a.
点评:考查了分式的化简求值,解答此题的关键是把分式化到最简,注意整体思想的应用.
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