谁帮我解这三个一元二次方程？

1.(2x-3)²=9(2x+3)²
(2x-3)²=(6x+9)²
(2x-3)²-(6x+9)²=0
(2x-3+6x+9)(2x-3-6x-9)=0
(8x+6)(-4x-12)=0
x=-3/4,x=-3

2.(5x-1)²=3（5x-1)
(5x-1)²-3(5x-1)=0
(5x-1)(5x-1-3)=0
(5x-1)(5x-4)=0
x=1/5,x=4/5

3.(x+1）²=-(x+1)+56
令a=x+1
则a²+a-56=0
(a-7)(a+8)=0
a=7,a=-8
x=a-1
所以 x=6,x=-9

1.（2x-3)²=9（2x+3)²
（2x-3)²-（6x+9)²=0
（2x-3+6x+9)*（2x-3-6x-9)=0
（4x-3)*（4x+12)=0
4x-3=0.....x=3/4
4x+12=0....x=-3
2.(5x-1)²=3（5x-1)
(5x-1)²-3（5x-1)=0
(5x-1)*（5x-4)=0
5x-1=0.....x=1/5
5x-4=0.....x=4/5
3.（x+1）²=-（x+1)+56
（x+1）²+（x+1)-56=0
（x+1-7）*（x+1+8)=0
（x-6）*（x+9)=0
x-6=0.....x=6
x+9=0.....x=-9