用因式分解法解下列方程: (1)(x+2)2=3x+6; (2)(3x+2)2-4x2=0; (3)5(2x-1)=(1-2x)(x+3); (4)2(x-3)2+(3x-x2)=0.
问题描述:
用因式分解法解下列方程:
(1)(x+2)2=3x+6;
(2)(3x+2)2-4x2=0;
(3)5(2x-1)=(1-2x)(x+3);
(4)2(x-3)2+(3x-x2)=0.
答
(1)原方程可变形为(x+2)(x+2-3)=0,(x+2)(x-1)=0.x+2=0或x-1=0.∴x1=-2,x2=1.(2)原方程可变形为(3x+2-2x)(3x+2+2x)=0,即(x+2)(5x+2)=0.x+2=0或5x+2=0.∴x1=-2,x2=-25.(3)原方程可变...