题目内容
先化简,再求值.(| 2 |
| x-2 |
| x-2 |
| x2-4x+4 |
| x2+2x |
| x-2 |
分析:先化简,再进行四则运算,再化简,把x2+2x-4=0,即x(x+2)=4代入,即可得值.
解答:解:∵x2+2x-4=0,
∴x(x+2)=4,
∵(
-
)÷
,
=(
-
)•
,
=
-
=
,
∴当x(x+2)=4时,原式=
=
.
∴x(x+2)=4,
∵(
| 2 |
| x-2 |
| x-2 |
| x2-4x+4 |
| x2+2x |
| x-2 |
=(
| 2 |
| x-2 |
| x-2 |
| (x-2)2 |
| x-2 |
| x(x-2) |
=
| 2 |
| x(x+2) |
| 1 |
| x(x+2) |
| 1 |
| x(x+2) |
∴当x(x+2)=4时,原式=
| 1 |
| x(x+2) |
| 1 |
| 4 |
点评:本题除考查了分式的混合运算外,还考查了分式方程的化简.解答此题的关键是把分式化到最简,然后代式计算.
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