Question

已知多项式x^2-2xy+y^2-x+y-1的值等于0,求x-y的值

Answer 1

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