用因式分解法解下列方程：1：y(y+5)=-25/4;2:3(x-3)=(x-3)^2;3:(2x+3)^2=24x4:1.2y-0.04=9y^2

1：y(y+5)=-25/4
y²+5y+25/4=0
(y+5/2)²=0
y=-5/2
2:3(x-3)=(x-3)^2;
(x-3)(3-x+3)=0
(x-3)(6-x)=0
x=3 x=6
3:(2x+3)^2=24x
4x²+12x+9-24x=0
4x²-12x+9=0
(2x-3)²=0
x=3/2
4:1.2y-0.04=9y^2
9y²-1.2y+0.04=0
(3y-0.2)²=0
y=0.2/3=1/15

1 y^2+5y+25/4=0
(y+5/2)^2=0
y+5/2=0
y=-5/2
2:3(x-3)=(x-3)^2;
3x-9=x^2-6x+9
3:(2x+3)^2=24x
4x²+12x+9-24x=0
4x²-12x+9=0
(2x-3)²=0
x=3/2
4:1.2y-0.04=9y^2
9y²-1.2y+0.04=0
(3y-0.2)²=0
y=0.2/3=1/15

1：y(y+5)=-25/4
y²+5y+25/4=0
(y+5/2)²=0
y=-5/2
2:3(x-3)=(x-3)^2;
(x-3)(3-x+3)=0
(x-3)(6-x)=0
x=3 x=6
3:(2x+3)^2=24x
4x²+12x+9-24x=0
4x²-12x+9=0
(2x-3)²=0
x=3/2
4:1.2y-0.04=9y^2
9y²-1.2y+0.04=0
(3y-0.2)²=0
y=0.2/3=1/15