题目内容
(1)
+
,其中x=-1;
(2)
÷(x-1-
),其中x=
.
| x+3 |
| x2-9 |
| 3 |
| x-3 |
(2)
| x2-2x |
| x2-1 |
| 2x-1 |
| x+1 |
| 1 |
| 2 |
分析:(1)先把原式统分化为最简形式,再把x=-1代入进行计算即可;
(2)先根据分式混合运算的法则把原式进行化简,再把x=
代入进行计算即可.
(2)先根据分式混合运算的法则把原式进行化简,再把x=
| 1 |
| 2 |
解答:解:(1)原式=
=
=
,
当x=-1时,原式=
=-1;
(2)原式=
÷(
)
=
÷
=
×
=
,
当x=
时,原式=
=
=-2.
| x+3+3(x+3) |
| x2-9 |
=
| 4(x+3) |
| x2-9 |
=
| 4 |
| x-3 |
当x=-1时,原式=
| 4 |
| -1-3 |
(2)原式=
| x2-2x |
| x2-1 |
| (x+1)(x-1)-2x+1 |
| x+1 |
=
| x2-2x |
| x2-1 |
| x(x-2) |
| x+1 |
=
| x(x -2) |
| x2-1 |
| x+1 |
| x(x-2) |
=
| 1 |
| x-1 |
当x=
| 1 |
| 2 |
| 1 |
| x-1 |
| 1 | ||
|
点评:本题考查的是分式的混合运算,熟知分式混合运算的法则是解答此题的关键.
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