题目

一元二次方程[因式分解]
3x^2+2x=0
3x^2-√5 x=0
x^2=(√3+1)x=0
x(x-1)+3(x-1)=0
3(x-5)^2=2(5-x)

提问时间：2020-10-08

答案

3x^2+2x=0
x(3x+2)=0
x=0,x=-2/3
3x^2-√5 x=0
x(3x-√5)=0
x=0,x=√5/3
x^2+(√3+1)x=0
x(x+√3+1)=0
x=0,x=-√3-1
x(x-1)+3(x-1)=0
(x-1)(x+3)=0
x=1,x=-3
3(x-5)^2=2(5-x)
3(x-5)^2+2(x-5)=0
(x-5)(3x-15+2)=0
(x-5)(3x-13)=0
x=5,x=13/3