问题补充：
xy(xy+1)+(xy+3)-2(x+y+2分之一）-(x+y-1)的平方进行因式分解
答案：
xy(xy+1)+(xy+3)-2(x+y+1/2)-(x+y-1)^2
=xy(xy+1)+(xy+1)+2-2(x+y)-1-[(x+y)^2-2(x+y)+1]
=(xy+1)^2+2-2(x+y)-1-(x+y)^2+2(x+y)-1
=(xy+1)^2-(x+y)^2
=(xy+x+y+1)(xy-x-y+1)
=(x+1)(y+1)(x-1)(y-1)
======以下答案可供参考======
供参考答案1:
原式=xy(xy+1)+{(xy+1)+2}-{2(x+y)+1}-{(x+y)-1}^2
=xy(xy+1)+(xy+1)+2-2(x+y)-1-{(x+y)^2-2(x+y)+1}
=(xy+1)(xy+1)-2(x+y)-(x+y)^2+2(x+y)
=(xy+1)^2-(x+y)^2
