题目内容
先化简,再求值:(1)x (x+2)-(x+1)(x-1),其中x=-
| 1 |
| 2 |
(2)(x+3)2+(x+2)(x-2)-2x2,其中x=-
| 1 |
| 3 |
(3)
| 1 |
| x+1 |
| 1 |
| x2-1 |
| x+1 |
| x2-2x+1 |
| 3 |
分析:(1)(2)首先利用整式的乘法法则去掉括号,然后化简,最后代入数值计算即可求解;
(3)首先把分式通分、约分,然后化简,最后代入数值计算即可求解.
(3)首先把分式通分、约分,然后化简,最后代入数值计算即可求解.
解答:解:(1)x (x+2)-(x+1)(x-1)
=x2+2x-x2+1
=2x+1
当x=-
时,原式=0;
(2)(x+3)2+(x+2)(x-2)-2x2
=x2+6x+9+x2-4-2x2
=6x+5,
当x=-
时,原式=3;
(3)
-
÷
=
-
×
=
,
当x=
-1时,原式=
.
=x2+2x-x2+1
=2x+1
当x=-
| 1 |
| 2 |
(2)(x+3)2+(x+2)(x-2)-2x2
=x2+6x+9+x2-4-2x2
=6x+5,
当x=-
| 1 |
| 3 |
(3)
| 1 |
| x+1 |
| 1 |
| x2-1 |
| x+1 |
| x2-2x+1 |
=
| 1 |
| x+1 |
| 1 |
| (x+1)(x-1) |
| (x-1)2 |
| x+1 |
=
| 2 |
| (x+1)2 |
当x=
| 3 |
| 2 |
| 3 |
点评:此题分别考查了整式的化简求值,分式的化简求值,解题的关键都是化简,然后代入数值计算即可求解.
练习册系列答案
相关题目