已知多项式x^2-2xy+y^2-x+y-1的值等于0,求x-y的值
问题描述：
已知多项式x^2-2xy+y^2-x+y-1的值等于0,求x-y的值
答
x^2-2xy+y^2-x+y-1
=（x-y）²-(x-y)-1
=0，
设(x-y)为z，则z²-z-1=0
解得：z1=½(1+√5)，z2=½(1-√5)
即：x-y=½(1+√5)，或x-y=½(1-√5)
答
x^2-2xy+y^2-x+y-1=0
(x-y)^2-(x-y)-1=0
令a=x-y
则a^2-a-1=0
a=(1±√5)/2
所以x-y=(1-√5)/2,x-y=(1+√5)/2