题目内容
已知| x2 |
| x2-2 |
| 1 | ||||
1-
|
| 1 |
| 1-x |
| 1 |
| 1+x |
| x |
| x2-1 |
分析:将分式通分、化简,再将已知条件变形,整体代入.
解答:解:(
-
)÷(
+x)
=-
÷
=-
•
=-
∵
=
,
∴
=1-
-
,
即-
=-(
+
)
∴原式=-(
+
).
| 1 |
| 1-x |
| 1 |
| 1+x |
| x |
| x2-1 |
=-
| 2x |
| (1-x)(1+x) |
| x3 |
| (1-x)(1+x) |
=-
| 2x |
| (1-x)(1+x) |
| (1-x)(1+x) |
| x3 |
| 2 |
| x2 |
∵
| x2 |
| x2-2 |
| 1 | ||||
1-
|
∴
| x2-2 |
| x2 |
| 2 |
| 3 |
即-
| 2 |
| x2 |
| 2 |
| 3 |
∴原式=-(
| 2 |
| 3 |
点评:本题主要考查分式的化简,整体代入的思想.
练习册系列答案
相关题目