题目内容
解方程.
(1)y2-2y-4=0(公式法);
(2)2x2-3x-5=0(配方法);
(3)(2x-1)2=9;
(4)(x-2)2=2x-4.
(1)y2-2y-4=0(公式法);
(2)2x2-3x-5=0(配方法);
(3)(2x-1)2=9;
(4)(x-2)2=2x-4.
(1)∵a=1,b=-2,c=-4,△=b2-4ac=4+16=20,
∴x=
,
解得x1=1+
,x2=1-
;
(2)系数化为1,得x2-
x-
=0,
配方得x2-
x+
-
-
=0,
即(x-
)2=
,
开方得,x-
=±
,
解得x1=
,x2=-1,
(3)直接开方得,2x-1=±3,
解得x1=2,x2=-1;
(4)(x-2)2=2x-4
移项,得(x-2)2-2(x-2)=0,
提公因式,得(x-2)(x-2-2)=0,
即x-2=0或x-4=0,
解得x1=2,x2=4.
∴x=
2±
| ||
| 2 |
解得x1=1+
| 5 |
| 5 |
(2)系数化为1,得x2-
| 3 |
| 2 |
| 5 |
| 2 |
配方得x2-
| 3 |
| 2 |
| 9 |
| 16 |
| 9 |
| 16 |
| 5 |
| 2 |
即(x-
| 3 |
| 4 |
| 49 |
| 16 |
开方得,x-
| 3 |
| 4 |
| 7 |
| 4 |
解得x1=
| 5 |
| 2 |
(3)直接开方得,2x-1=±3,
解得x1=2,x2=-1;
(4)(x-2)2=2x-4
移项,得(x-2)2-2(x-2)=0,
提公因式,得(x-2)(x-2-2)=0,
即x-2=0或x-4=0,
解得x1=2,x2=4.
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