题目内容：

已知x=3的平方根+1,y=3的平方根-1,求x^2-2xy+y^2/x^2
求x^2-2xy+y^2/x^2-y^2的值

优质解答

已知x=±√3+1,y=±√3-1
(x²-2xy+y²)/(x²-y²)
=(x-y)²/(x+y)(x-y)
=(x-y)/(x+y)
前面知道分母x²-y²≠0
所以x²≠y²
x≠y,x≠-y
所以x=√3+1,y=√3-1或x=-√3+1,y=-√3-1
①
x=√3+1,y=√3-1时
原式=(x-y)/(x+y)
=(√3+1-√3+1)/(√3+1+√3-1)
=2/2√3
=√3/3
②
x=-√3+1,y=-√3-1时
原式=(x-y)/(x+y)
=(-√3+1+√3+1)/(-√3+1-√3-1)
=2/(-2√3)=-√3/3