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

2025-04-29 07:00:40
推荐回答(1个)
回答1:

(1)x2-5x+1=0,
移项得:x2-5x=-1,
配方得:x2-5x+

25
4
=-1+
25
4

即(x-
5
2
2=
21
4

∴x-
5
2
21
2

∴x1=
5+
21
2
,x2=
5?
21
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
x?5=0

∵a=2,b=-2
2
,c=-5,
∴△=8-4×2×(-5)=48,
∴x=
2
2
±