(2x+3)²-4(2x+3)+4=0 (x-3)²=(3x-2)²一元二次方程分解因式法怎么解过程
问题描述:
(2x+3)²-4(2x+3)+4=0 (x-3)²=(3x-2)²一元二次方程分解因式法怎么解过程
答
(2x+3)²-4(2x+3)+4=0 [(2x+3)-2]^2=02x+3-2=02x=-1x=-1/2.(x-3)²=(3x-2)²(x-3)^2-(3x-2)^2=0(x-3+3x-2)(x-3-3x+2)=0(4x-5)(-2x-1)=0x1=5/4x2=-1/2