题目内容
化简| x2+1 |
| x2-1 |
| x-2 |
| x-1 |
| x-2 |
| 2 |
分析:首先把除法运算转化成乘法运算,进行约分,然后进行减法运算,进行通分化简,最后代值计算.
解答:解:原式=
-
•
=
-
=
-
=
=
=-
;
当x=5时,原式=-
=-
=-
.
| x2+1 |
| (x+1)(x-1) |
| x-2 |
| x-1 |
| x |
| x-2 |
| x2+1 |
| (x+1)(x-1) |
| x |
| x-1 |
=
| x2+1 |
| (x+1)(x-1) |
| x(x+1) |
| (x+1)(x-1) |
| x2+1-x2-x |
| (x+1)(x-1) |
| 1-x |
| (x+1)(x-1) |
| 1 |
| x+1 |
当x=5时,原式=-
| 1 |
| x+1 |
| 1 |
| 5+1 |
| 1 |
| 6 |
点评:本题主要考查分式的化简求值,把分式化到最简是解题的关键.
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