把多项式（xy-1）²+（x+y-2）（x+y-2xy）分解因式
问题描述：
把多项式（xy-1）²+（x+y-2）（x+y-2xy）分解因式
答
(xy-1)^2+(x+y-2)(x+y-2xy)
=x^2y^2-2xy+1+(x+y)^2-2(x+y)-2xy(x+y)+4xy
=x^2y^2+2xy+1+(x+y)^2-2(x+y)(xy+1)
=(xy+1)^2-2(x+y)(xy+1)+(x+y)^2
=(xy+1-x-y)^2
=(x-1)^2(y-1)^2
答
（xy-1）²+（x+y-2）（x+y-2xy）
=x²y²-2xy+1+(x+y)²-2xy(x+y)-2(x+y)+4xy
=x²y²+2xy+1+(x+y)²-2xy(x+y)-2(x+y)
=(xy+1)²-2(xy+1)(x+y)+(x+y)²
=(xy+1-x-y)²
=[x(y-1)-(y-1)]²
=(x-1)²(y-1)²