由x^2-2xy+y^2-x+y-1=0,求x-y=

(x-y)^2-(x-y)-1=0
令x-y=m
m^2-m-1=0
解出方程值就行了
x-y=
1/2*5^(1/2)+1/2
-1/2*5^(1/2)+1/2

(1+/-根号5）/2

(X-Y)^2-(x-y)-1=0
这样会做了吧

(x-y)^2-(x-y)-1=0
(x-y)^2-(x-y)=1
提公因式(x-y)(x-y-x+y)=1
x-y=1

原题可变为：
（x-y)^2-(x-y)-1=0
[(x-y)-1/2]^2-1/4-1=0
[(x-y)-1/2]^2=5/4
(x-y)-1/2=正负根号5/2
解得
（x-y)=1/2+根号5／2或者(x-y)=1/2-根号5／2