题目内容
(2011•南岗区二模)先化简,再求值:
÷ (x-1-
),其中x=2cos45°+1.
| x2-2x |
| x2-1 |
| 2x-1 |
| x+1 |
分析:先把括号内通分得到原式=
÷
,然后把各分式的分子或分母因式分解得原式=
•
,约分后得原式=
,然后把x值代入计算即可.
| x(x-2) |
| (x+1)(x+1) |
| (x-1)(x+1)-(2x-1) |
| x+1 |
| x(x-2) |
| (x+1)(x+1) |
| x+1 |
| x(x-2) |
| 1 |
| x-1 |
解答:解:原式=
÷
=
÷
=
•
=
,
当x=2cos45°+1=2×
+1=
+1,
原式=
=
=
.
| x(x-2) |
| (x+1)(x+1) |
| (x-1)(x+1)-(2x-1) |
| x+1 |
=
| x(x-2) |
| (x+1)(x+1) |
| x2-1-2x+1 |
| x+1 |
=
| x(x-2) |
| (x+1)(x+1) |
| x+1 |
| x(x-2) |
=
| 1 |
| x-1 |
当x=2cos45°+1=2×
| ||
| 2 |
| 2 |
原式=
| 1 | ||
|
| 1 | ||
|
| ||
| 2 |
点评:本题考查了分式的化简求值:先把括号内通分,再把各分式的分子或分母因式分解,然后约分得到最简分式或整式,再把满足条件的字母的值代入计算即可.
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