用分解因式解下列方程1.(4x-1)(5x+7)=0 2.3x(x-1)=2-2x3.(2x+3)^2=4(2x+3)4

(4x-1)(5x+7)=0
x1=1/4 x2=-7/5

3x(x-1)=2-2x
3x^2-3x=2-2x
3x^2-x-2=0
(3x-2)(x+1)=0
x=2/3 x=-1
(2x+3)^2=4(2x+3)
(2x+3)^2-4(2x+3)=0
(2x+3)(2x+3-4)=0
(2x+3)(2x-1)=0
x=-3/2 x=1/2
2(x-3)^2=x^2-9
2(x-3)^2-(x^2-9)=0
2(x-3)^2-(x+3)(x-3)=0
(x-3)(2(x-3)-(x+3))=0
(x-3)(2x-6-x-3)=0
(x-3)(x-9)=0
x=3 x=9

(1) (4x-1)=0或(5x+7)=0
x1= 1/4 x2= -7/5
未完，待续，先加分！

1.(4x-1)(5x+7)=0 ＝> x=1/4 x=-7/5
2.3x(x-1)=2-2x =>(x-1)(3x+2)=0 =>x=1 x=-2/3
3.(2x+3)^2=4(2x+3) =>(2x+3)(2x+3-4)=0 =>x=-3/2 x=1/2
4.2(x-3)^2=x^2-9 => (x-3)(2x-6-x-3)=0=>x=3 x=9

2: (x-1)(3x-2)=0
3:(2x+3)(2x-1)=0
4:(x-3)(x-9)=0

1.(4x-1)(5x+7)=0 或
4x-1=0 或 5x+7=0
x=1/4 或 x=-7/5
2.3x(x-1)=2-2x
3x(x-1)=-2(x-1)
(x-1)(3x+2)=0
x=1或x=-2/3
3.(2x+3)^2=4(2x+3)
(2x+3)(2x+3-4)=0
(2x+3)(2x-1)=0
x=-3/2或x=1/2
4.2(x-3)^2=x^2-9
2(x-3)^2-(x+3)(x-3)=0
(x-3)[2(x-3)-(x+3)]=0
(x-3)(x-9)=0
x=3或x=9