xy(xy+1)+(xy+3)-2(x+y+2分之一）-(x+y-1)的平方
进行因式分解
參考答案: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)
xy(xy+1)+(xy+3)-2(x+y+2分之一）-(x+y-1)的平方
进行因式分解
參考答案: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)