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=0a=(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)
