一元二次方程的因式分解法3(x-5)^2+=2(x-5)和(3x-4)^2=9x-12和(2x+3)^2-4(2x+3)
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解析:

3(x-5)^2=2(x-5)

3(x-5)^2-2(x-5)=0,

(x-5)[3(x-5)-2]=0

(x-5)(3x-17)=0,

∴x1=5,x2=17/3,

(3x-4)^2=9x-12,

(3x-4)^2-(9x-12)=0

(3x-4)^2-3(x-4)=0,

(3x-4)[(3x-4)-3]=0

(3x-4)(3x-7)=0

∴x1=4/3,x2=7/3

(2x+3)^2-4(2x+3)+4=0

[(2x+3)-4]^2=0,

即(2x-1)^2=0

∴x1=x2=1/2