题目内容
先化简,再求值:(| x+2 |
| x2-2x |
| x-1 |
| x2-4x+4 |
| x2-16 |
| x2+4x |
| 3 |
分析:原式括号内各式乘以x(x-2)2再根据约分的方法和二次根式的性质进行化简,最后代入计算.
解答:解:原式=[
-
]÷
=[
-
]÷
=
•
=
,
当x=2+
时,
原式=
=
.
| x+2 |
| x(x-2) |
| x-1 |
| (x-2)2 |
| x2-16 |
| x2+4x |
=[
| x2-4 |
| x(x-2)2 |
| x2-x |
| x(x-2)2 |
| x2-16 |
| x2+4x |
=
| x-4 |
| x(x-2)2 |
| x(x+4) |
| (x+4)(x-4) |
=
| 1 |
| (x-2)2 |
当x=2+
| 3 |
原式=
| 1 | ||
(2+
|
=
| 1 |
| 3 |
点评:本题主要考查二次根式的化简求值,化简过程一定要细心.
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