题目内容
已知:x=| 2 |
| x2-4 |
| x |
| x+2 |
| x2-2x |
| x-1 |
| x2-4x+4 |
| x-4 |
| x |
分析:首先把括号里的式子进行通分,然后把除法运算转化成乘法运算,进行约分化简,最后代值计算.
解答:解:原式=
•(
-
)•
=
•
•
=
•
•
=
;
当x=
时,原式=
=
=
(或-2-
).
| x2-4 |
| x |
| x+2 |
| x(x-2) |
| x-1 |
| (x-2)2 |
| x |
| x-4 |
=
| x2-4 |
| x |
| x2-4-x2+x |
| x(x-2)2 |
| x |
| x-4 |
=
| x2-4 |
| x |
| x-4 |
| x(x-2)2 |
| x |
| x-4 |
=
| x+2 |
| x(x-2) |
当x=
| 2 |
| ||||
|
6+4
| ||
-2
|
4+3
| ||
| 2 |
3
| ||
| 2 |
点评:这是典型的“化简求值”的题目,着眼于对运算法则的掌握和运算能力的直接考查,有着很好的基础性和效度.这是个分式混合运算题,运算顺序是先乘除后加减,加减法时要注意把各分母先因式分解,确定最简公分母进行通分.
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