题目内容
先化简,再求值:| x2-x |
| x+1 |
| 2x-2 |
| x+1 |
| 2 |
分析:先计算括号里的减法,再计算除法,最后代值计算.
解答:解:原式=
÷
=
•
=
;
当x=
+1时,
原式=
=
=
.
| x(x-1) |
| x+1 |
| (x2-1-2x+2) |
| x+1 |
=
| x(x-1) |
| x+1 |
| x+1 |
| (x-1)2 |
=
| x |
| x-1 |
当x=
| 2 |
原式=
| ||
|
| ||
|
2+
| ||
| 2 |
点评:此题是一道基础题,学生只要能认真化简代值,就能正确地解此题.
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