题目内容
先化简,再求值
(1)
-
(其中x=
)
(2)(
-
)÷
(其中x=
-1)
(1)
| 1-2x+x2 |
| x-1 |
| ||
| x2-x |
| 1 | ||
2+
|
(2)(
| x |
| x-1 |
| ||
| x+1 |
| 1 |
| x2-1 |
| 3 |
分析:(1)先求出x,再求出每一部分的值,合并后代入求出即可;
(2)把除法变成乘法,根据多项式乘以单项式法则展开,合并后代入求出即可.
(2)把除法变成乘法,根据多项式乘以单项式法则展开,合并后代入求出即可.
解答:解:(1)∵x=
=
=2-
<1,
∴原式=
-
=x-1-
=x-1+x
=2x-1
=2(2-
)-1
=3-2
.
(2)(
-
)÷
=(
-
)×
=x(x+1)-
(x-1)
∵x=
-1,
∴=(
-1)×(
-1+1)-
(
-1-1)
=3-
-3+2
=
.
| 1 | ||
2+
|
1×(2-
| ||||
(2+
|
| 3 |
∴原式=
| (x-1)2 |
| x-1 |
| ||
| x(x-1) |
=x-1-
| 1-x |
| x(x-1) |
=x-1+x
=2x-1
=2(2-
| 3 |
=3-2
| 3 |
(2)(
| x |
| x-1 |
| ||
| x+1 |
| 1 |
| x2-1 |
=(
| x |
| x-1 |
| ||
| x+1 |
| (x+1)(x-1) |
| 1 |
=x(x+1)-
| 3 |
∵x=
| 3 |
∴=(
| 3 |
| 3 |
| 3 |
| 3 |
=3-
| 3 |
| 3 |
=
| 3 |
点评:本题考查了二次根式和分式的化简求值的应用,主要考查学生的化简和计算能力.
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