题目内容
先化简,再求值:
(1)(
-
)÷
,其中x=1.
(2)
-
÷
,其中x=
-1.
(1)(
| 1 |
| x2-2x |
| 1 |
| x2-4x+4 |
| 2 |
| x2-2x |
(2)
| 1 |
| x+1 |
| 1 |
| x2-1 |
| x+1 |
| x2-2x+1 |
| 3 |
考点:分式的化简求值
专题:
分析:(1)、(2)先根据分式混合运算的法则把原式进行化简,再把x的值代入进行计算即可.
解答:解:(1)原式=(
-
)÷
=
•
=
•
=
,
当x=1时,原式=
=1;
(2)原式=
-
•
=
-
=
=
,
当x=
-1时,原式=
=
.
| 1 |
| x(x-2) |
| 1 |
| (x-2)2 |
| 2 |
| x(x-2) |
=
| x-2-x |
| x(x-2)2 |
| x(x-2) |
| 2 |
=
| -2 |
| x(x-2)2 |
| x(x-2) |
| 2 |
=
| -1 |
| x-2 |
当x=1时,原式=
| -1 |
| 1-2 |
(2)原式=
| 1 |
| x+1 |
| 1 |
| (x+1)(x-1) |
| (x-1)2 |
| x+1 |
=
| 1 |
| x+1 |
| x-1 |
| (x+1)2 |
=
| x+1-x+1 |
| (x+1)2 |
=
| 2 |
| (x+1)2 |
当x=
| 3 |
| 2 | ||
(
|
| 2 |
| 3 |
点评:本题考查的是分式的化简求值,熟知分式混合运算的法则是解答此题的关键.
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