题目内容
解方程:
①2x2-5x-1=0(限用配方法);
②3(x-2)2=x(x-2)(限用因式分解法).
①2x2-5x-1=0(限用配方法);
②3(x-2)2=x(x-2)(限用因式分解法).
①原方程化为2x2-5x=1,
x2-
x=
,
x2-
x+(
)2=
+(
)2,
(x-
)2=
,即x-
=±
,
x1=
+
,x2=
-
.
②原方程化为
3(x-2)2-x(x-2)=0
(x-2)(3x-6-x)=0
x1=2,x2=3.
x2-
| 5 |
| 2 |
| 1 |
| 2 |
x2-
| 5 |
| 2 |
| 5 |
| 4 |
| 1 |
| 2 |
| 5 |
| 4 |
(x-
| 5 |
| 4 |
| 33 |
| 16 |
| 5 |
| 4 |
| ||
| 4 |
x1=
| 5 |
| 4 |
| ||
| 4 |
| 5 |
| 4 |
| ||
| 4 |
②原方程化为
3(x-2)2-x(x-2)=0
(x-2)(3x-6-x)=0
x1=2,x2=3.
练习册系列答案
相关题目