题目内容
(2013•莒南县一模)先化简,再求值:1÷(
-
)×
,其中x的值是方程x2+x-6=0的根.
| x+2 |
| x2-2x |
| x-1 |
| x2-4x+4 |
| x+2 |
| x3-4x |
分析:先计算出括号内的和,再因式分解,然后将除法转化为乘法,再将x2+x-6=0解出来,把x的值代入即可求解.
解答:解:原式=1÷[
-
]×
=1÷[
-
]×
=1÷
×
=1÷
×
=1×
×
=
,
由x2+x-6=0解得x1=-3,x2=2(舍去),
当x=-3时,原式=
.
当x=2时,原式时无意义.
| x+2 |
| x(x-2) |
| x-1 |
| (x-2)2 |
| x+2 |
| x(x-2)(x+2) |
=1÷[
| x2-4 |
| x(x-2)2 |
| x2-x |
| x(x-2)2 |
| x+2 |
| x(x-2)(x+2) |
=1÷
| x-4 |
| x(x-2)2 |
| x+2 |
| x(x-2)(x+2) |
=1÷
| x-4 |
| x(x-2)2 |
| 1 |
| x(x-2) |
=1×
| x(x-2)2 |
| x-4 |
| 1 |
| x(x-2) |
=
| x-2 |
| x-4 |
由x2+x-6=0解得x1=-3,x2=2(舍去),
当x=-3时,原式=
| 5 |
| 7 |
当x=2时,原式时无意义.
点评:本题考查了分式的化简求值,涉及一元二次方程的解法,要熟练掌握因式分解.
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