题目内容
先化简,再求值:
-
÷(x+1-
),其中x满足x2+2x-4=0.
| 1 |
| x+2 |
| x2-4x+4 |
| x2-x |
| 3 |
| x-1 |
考点:有理数指数幂的化简求值
专题:计算题,函数的性质及应用
分析:化简
-
÷(x+1-
)=
-
×
=
-
=
,再代入求值.
| 1 |
| x+2 |
| x2-4x+4 |
| x2-x |
| 3 |
| x-1 |
| 1 |
| x+2 |
| (x-2)2 |
| x(x-1) |
| x-1 |
| (x-2)(x+2) |
| 1 |
| x+2 |
| x-2 |
| x(x+2) |
| 2 |
| x(x+2) |
解答:
解:
-
÷(x+1-
)
=
-
×
=
-
=
,
∵x2+2x-4=0,
∴x(x+2)=4;
故
=
;
即
-
÷(x+1-
)=
.
| 1 |
| x+2 |
| x2-4x+4 |
| x2-x |
| 3 |
| x-1 |
=
| 1 |
| x+2 |
| (x-2)2 |
| x(x-1) |
| x-1 |
| (x-2)(x+2) |
=
| 1 |
| x+2 |
| x-2 |
| x(x+2) |
=
| 2 |
| x(x+2) |
∵x2+2x-4=0,
∴x(x+2)=4;
故
| 2 |
| x(x+2) |
| 1 |
| 2 |
即
| 1 |
| x+2 |
| x2-4x+4 |
| x2-x |
| 3 |
| x-1 |
| 1 |
| 2 |
点评:本题考查了表达式的化简与运算,属于基础题.
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