用因式分解法解下列方程：1：y(y+5)=-25/4;

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

原式[2(y-3)]²-[5(y-2)]²=0 ==>(2y-6)²-(5y-10)²=0 ==>(2y-6-5y+10)(2y-6+5y-10)=0 ==>(-3y+4)(7y-16)=0 y1=4/3  y2=16/7

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 4x +12x+9-24x=0 4x -12x+9=0 (2x-3) =0 x=3/2 4:1.2y-0.04=9y^2 9y -1.2y+0.04=0 (3y-0.2) =0 y=0.2/3=1/15