设x,y为正数,且满足x²-2xy-y²=0,求x-y\x+y的值.

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设x,y为正数,且满足x²-2xy-y²=0,求x-y\x+y的值.
若分解x²-x+m得到两个因式x-2与x-n，求m与n的值。

$设x,y为正数,且满足x²-2xy-y²=0,求x-y\x+y的值.$

x^2-2xy-y^2=0
x^2-2xy+y^2=2y^2
(x-y)^2=2y^2
|x-y|=根号2 Y
二边同除以Y得到:|X/Y-1|=根号2
即X/Y=1(+/-)根号2
(X-Y)/(X+Y)=(X/Y-1)/(X/Y+1)
(1)X/Y=1+根号2,原式=(1+根号2-1)/(1+根号2+1)=根号2/(2+根号2)=根号2*(2-根号2)/2=根号2-1
(2)X/Y=1-根号2,原式=(1-根号2-1)/(1-根号2+1)=[-根号2]/(2-根号2)=-根号2(2+根号2)/2=-根号2-1
再问：若分解x²-x+m得到两个因式x-2与x-n，求m与n的值。
再答： x^2-x+m=(x-2)(x-n)=x^2-(2+n)x+2n 2+n=1 m=2n n=-1,m=-2

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