题目内容
解方程:
(1)(x+1)2-5=0;
(2)2x2-4x+1=0;
(3)(2x+1)(x-3)=-7;
(4)3(x-2)2=x(x-2).
(1)(x+1)2-5=0;
(2)2x2-4x+1=0;
(3)(2x+1)(x-3)=-7;
(4)3(x-2)2=x(x-2).
(1)(x+1)2=5,
x+1=±
,
∴x1=
-1,x2=-
-1;
(2)∵a=2,b=-4,c=1,
∴x=
=
,
即x1=
,x2=
;
(3)整理,得2x2-5x+4=0,
∵△=25-32=-7<0,
∴原方程无实数根;
(4)3(x-2)2-x(x-2)=0,
(x-2)[3(x-2)-x]=0,
∴x-2=0或3(x-2)-x=0,
∴x1=2,x2=3.
x+1=±
| 5 |
∴x1=
| 5 |
| 5 |
(2)∵a=2,b=-4,c=1,
∴x=
-b±
| ||
| 2a |
2±
| ||
| 2 |
即x1=
2+
| ||
| 2 |
2-
| ||
| 2 |
(3)整理,得2x2-5x+4=0,
∵△=25-32=-7<0,
∴原方程无实数根;
(4)3(x-2)2-x(x-2)=0,
(x-2)[3(x-2)-x]=0,
∴x-2=0或3(x-2)-x=0,
∴x1=2,x2=3.
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