题目内容
| u-2v |
| u+2v |
| 2 |
| u2-4v2 |
分析:首先通分,将原式通分为
-
,然后利用同分母的分式相加减的运算法则求解即可求得答案.
| (u-2v)2 |
| (u+2v)(u-2v) |
| 2 |
| (u+2v)(u-2v) |
解答:解:
-
=
-
=
.
| u-2v |
| u+2v |
| 2 |
| u2-4v2 |
| (u-2v)2 |
| (u+2v)(u-2v) |
| 2 |
| (u+2v)(u-2v) |
| u2-4uv+v2-2 |
| (u+2v)(u-2v) |
点评:此题考查了异分母分式的加减运算法则.注意先通分,再利用同分母的分式相加减的运算法则求解即可,注意运算结果需化为最简.
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