题目内容
先化简,再求值:(
-
)÷
-
,其中x满足2x2-2x-7=0.
| x+1 |
| x-2 |
| x-2 |
| x-1 |
| 4x2-5x |
| x2-4 |
| 1 |
| x-1 |
分析:先把括号里面的进行通分,再把除法转化成乘法,再进行约分,然后把2x2-2x-7=0进行整理,最后代入即可求出答案;
解答:解:原式=
•
-
=
•
-
,
=
-
,
=
,
∵2x2-2x-7=0,
∴x2-x=
,
∴原式=
;
| (x+1)(x-1)-(x-2)2 |
| (x-2)(x-1) |
| (x+2)(x-2) |
| x(4x-5) |
| 1 |
| x-1 |
=
| 4x-5 |
| x-1 |
| x+2 |
| x(4x-5) |
| 1 |
| x-1 |
=
| x+2 |
| x(x-1) |
| x |
| x(x-1) |
=
| 2 |
| x2-x |
∵2x2-2x-7=0,
∴x2-x=
| 7 |
| 2 |
∴原式=
| 4 |
| 7 |
点评:第一题考查的是分式的化简求值,特别注意运算顺序及符号的处理,也需要对通分、分解因式、约分等知识点熟练掌握.
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