已知多项式x^2-2xy+y^2-x+y-1的值等于0,求x-y的值

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已知多项式x^2-2xy+y^2-x+y-1的值等于0,求x-y的值

已知多项式x^2-2xy+y^2-x+y-1的值等于0,求x-y的值
已知多项式x^2-2xy+y^2-x+y-1的值等于0,求x-y的值

已知多项式x^2-2xy+y^2-x+y-1的值等于0,求x-y的值
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

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)