题目内容
已知| x |
| x-1 |
| y2+4y-2 |
| y2+4y-1 |
分析:此题可先从
=
下手,通过变形可得1+
=1-
,再变形即可求得结果.
| x |
| x-1 |
| y2+4y-2 |
| y2+4y-1 |
| 1 |
| x-1 |
| 1 |
| y2+4y-1 |
解答:解:由于
=
,则通过变形可得:1+
=1-
,
即
=-
,∴y2+4y+x=2.
| x |
| x-1 |
| y2+4y-2 |
| y2+4y-1 |
| 1 |
| x-1 |
| 1 |
| y2+4y-1 |
即
| 1 |
| x-1 |
| 1 |
| y2+4y-1 |
点评:本题考查了分式的化简求值,关键是从题中所给的等式下手,找到切入点.
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