问题
填空题
分解因式：x2y+xy2-x2-y2-3xy+2x+2y-1=______．
答案
x2y+xy2-x2-y2-3xy+2x+2y-1
=x2y+xy2-xy-x2-y2-2xy+2x+2y-1
=xy（x+y-1）-[（x+y）2-2（x+y）+1]
=xy（x+y-1）-（x+y-1）2
=（x+y-1）（xy-x-y+1）
=（x+y-1）（y-1）（x-1）．
故答案为：（x+y-1）（y-1）（x-1）．