题目内容
已知a=2-
,先化简,再求值:
-
-
.
| 3 |
| a2-1 |
| a+1 |
| ||
| a2-a |
| 1 |
| a |
分析:求出a-1的值,化简每一部分,再进行约分,最后合并,把a-1的值代入求出即可.
解答:解:∵a=2-
,
∴a-1=1-
∴
-
-
=
-
-
=a-1-
-
=a-1+
-
=a-1
=1-
.
| 3 |
∴a-1=1-
| 3 |
∴
| a2-1 |
| a+1 |
| ||
| a2-a |
| 1 |
| a |
=
| (a+1)(a-1) |
| a+1 |
| |a-1| |
| a(a-1) |
| 1 |
| a |
=a-1-
| -(a-1) |
| a(a-1) |
| 1 |
| a |
=a-1+
| 1 |
| a |
| 1 |
| a |
=a-1
=1-
| 3 |
点评:本题考查了二次根式的性质,二次根式的混合运算的应用,主要考查学生的化简能力和计算能力.
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