题目内容
化简求值:| x2+2x+1 |
| x+2 |
| x2-1 |
| x-1 |
| 1 |
| x+2 |
| 2 |
分析:首先把除法运算转化成乘法运算,进行因式分解,约分,然后进行减法运算,最后代值计算.
解答:解:原式=
÷
-
=
×
-
=
-
=
;
当x=时,原式=
=
=
-1.
| (x+1)2 |
| x+2 |
| (x+1)(x-1) |
| x-1 |
| 1 |
| x+2 |
=
| (x+1)2 |
| x+2 |
| 1 |
| x+1 |
| 1 |
| x+2 |
=
| x+1 |
| x+2 |
| 1 |
| x+2 |
=
| x |
| x+2 |
当x=时,原式=
| ||
|
| ||||
(2+
|
| 2 |
点评:此题主要考查分式的运算及求值.在分式的运算中遇到分子、分母是多项式时,一般先分解因式,再参与运算.
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