题目内容
已知x=
+1,则
÷
+
=
| 2 |
| x-2 |
| x2-1 |
| 2x+2 |
| x2+2x+1 |
| 1 |
| x-1 |
4
| 2 |
4
.| 2 |
分析:先把要求的式子进行因式分解,再把除法转化成乘法,再进行约分,最后把x的值代入即可求出答案.
解答:解:
÷
+
=
×
+
=
+
=
+
=
=
,
把x=
+1代入上式得:
原式=
=4
,
故答案为:4
.
| x-2 |
| x2-1 |
| 2x+2 |
| x2+2x+1 |
| 1 |
| x-1 |
=
| x-2 |
| (x+1)(x-1) |
| (x+1) 2 |
| 2(x+1) |
| 1 |
| x+1 |
=
| x-2 |
| 2(x-1) |
| 2(x-1) |
| 2(x+1)(x-1) |
=
| (x-2)(x-1) |
| 2(x-1)(x+1) |
| 2(x-1) |
| 2(x+1)(x-1) |
=
| (x-2)(x-1)+2(x-1) |
| 2(x-1)(x+1) |
=
| x |
| 2(x+1) |
把x=
| 2 |
原式=
| ||
2(
|
| 2 |
故答案为:4
| 2 |
点评:本题除考查了分式的化简求值,运算时要注意分子、分母能因式分解的先因式分解;除法要统一为乘法运算.还考查了分式方程的解.
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