Question

用分解因式法解下列方程3m²-4m=0（3x-4)²=（4x-3)²2x²-6x？

Answer 1

3m²-4m=0,m(3m-4)=0, m1=0；m2=4/3
(3x-4)²=(4x-3)²,(3x-4)²-(4x-3)²=0,[(3x-4)+(4x-3)]*[(3x-4)-(4x-3)]=0,
[(7x-7)]*[(-x-1)]=0, x1=1; x2=-1
2x²-6x=5x-15,2x²-11x+15=0,(2x-5)(x-3)=0,x1=5/2；x2=3
(2x-1)(x+1)=(3x+1)(x+1),(3x+1)(x+1)-(2x-1)(x+1)=0,(x+1)(3x+1-2x+1)=0,
(x+1)(x+2)=0,x1=-1；x2=-2
解方程
（x-3)²+2×（x-3）=0
(x-3)(x-3+2x)=0
(x-3)(3x-3)=0
x1=3
x2=1

2(x-3)=3x(x-3)
3x(x-3)-2(x-3)=0
(x-3)(3x-2)=0
x1=3
x2=2/3

x(x-2)+x-2=0
x(x-2)+(x-2)=0
(x-2)(x+1)=0
x1=2
x2=-1,9,用分解因式法解下列方程
3m²-4m=0
（3x-4)²=（4x-3)²
2x²-6x=5x-15
(2x-1)(x+1)=(3x+1)(x+1)
解方程
（x-3)²+2×（x-3）=0
2(x-3)=3x(x-3)
x(x-2)+x-2=0