解下列方程:(1)x2-5x+1=0(用配方法)(2)3(x-2)2=x(x-2)(3)2x2?22x?5=0(4)(y+2)2=(3y-1
解下列方程:(1)x2-5x+1=0(用配方法)(2)3(x-2)2=x(x-2)(3)2x2?22x?5=0(4)(y+2)2=(3y-1)2....
解下列方程:(1)x2-5x+1=0(用配方法)(2)3(x-2)2=x(x-2)(3)2x2?22x?5=0(4)(y+2)2=(3y-1)2.
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(1)x2-5x+1=0,
移项得:x2-5x=-1,
配方得:x2-5x+
=-1+
,
即(x-
)2=
,
∴x-
=±
,
∴x1=
,x2=
;
(2)3(x-2)2=x(x-2),
移项,得 3(x-2)2-x(x-2)=0,
(x-2)(3x-6-x)=0,
x-2=0或2x-6=0,
x1=2,x2=3;
(3)2x2?2
x?5=0,
∵a=2,b=-2
,c=-5,
∴△=8-4×2×(-5)=48,
∴x=
移项得:x2-5x=-1,
配方得:x2-5x+
25 |
4 |
25 |
4 |
即(x-
5 |
2 |
21 |
4 |
∴x-
5 |
2 |
| ||
2 |
∴x1=
5+
| ||
2 |
5?
| ||
2 |
(2)3(x-2)2=x(x-2),
移项,得 3(x-2)2-x(x-2)=0,
(x-2)(3x-6-x)=0,
x-2=0或2x-6=0,
x1=2,x2=3;
(3)2x2?2
2 |
∵a=2,b=-2
2 |
∴△=8-4×2×(-5)=48,
∴x=
2
|