题目内容
(2013•太仓市二模)先化简,再求值:
÷
-
,其中x=
.
| x-2 |
| x2-1 |
| x+1 |
| x2+2x+1 |
| x |
| x-1 |
| 1 | ||
|
分析:原式第一项利用除以一个数等于乘以这个数的倒数将除法运算化为乘法运算,约分后利用同分母分式的减法法则计算得到最简结果,将x的值代入计算即可求出值.
解答:解:原式=
÷
-
=
•
-
=
-
=-
,
当x=
=
+1时,原式=-
=-
.
| x-2 |
| (x+1)(x-1) |
| x+1 |
| (x+1)2 |
| x |
| x-1 |
=
| x-2 |
| (x+1)(x-1) |
| (x+1)2 |
| x+1 |
| x |
| x-1 |
=
| x-2 |
| x-1 |
| x |
| x-1 |
=-
| 2 |
| x-1 |
当x=
| 1 | ||
|
| 2 |
| 2 | ||
|
| 2 |
点评:此题考查了分式的化简求值,分式的加减运算关键是通分,通分的关键是找最简公分母;分式的乘除运算关键是约分,约分的关键是找公因式.
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