(xy-1)的平方-(x+y-2xy)(2-x-y)
问题描述：
(xy-1)的平方-(x+y-2xy)(2-x-y)
答
（xy-1)²-(x+y-2xy)(2-x-y)
=（xy-1)²+(x+y-2xy)(x+y-2)
=x²y²-2xy+1+(x+y)²-2(x+y)-2xy(x+y)+4xy
=x²y²-2x²y-2xy²+x²+y²+4xy-2x-2y+1不是因式分解吗？？？？因式分解因式分解最后不应该有那么多的多项式呀（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