Question

设x,y为正数,且满足x2-2xy-y2=0,求x-y/x+y的值？

Answer 1

x,y为正数,且满足x2-2xy-y2=0,(x-y)2=2y2所以x-y/x+y=(x-y)2/(x2-y2)=2y2/2xy=y/x由(x-y)2=2y2得x=(根号2+1)y或x=(-根号2+1)y
x-y/x+y=y/x=根号2-1或
x-y/x+y=y/x=-根号2-1

Answer 2

由x2-2xy-y2=0，（x-y）^2=2y^2,即x-y=根号2y，故x=（1+根号2）y，知道x与y的关系即可求x-y/x+y=根号2/(2+根号2)；

Answer 3

由方程解得x=(1+√2)y或x=(1-√2)y
当x=(1+√2)y时，代入方程得 √2/(2+√2)
当x=(1-√2)y时，代入方程得 √2/(√2-2)

Answer 4

x,y为正数,且满足x2-2xy-y2=0,(x-y)2=2y2所以x-y/x+y=(x-y)2/(x2-y2)=2y2/2xy=y/x由(x-y)2=2y2得x=(根号2+1)y或x=(-根号2+1)y
x-y/x+y=y/x=根号2-1或
x-y/x+y=y/x=-根号2-1 ∵x,y都为正数，第二个舍去 =根号2-1
　　　　

Answer 5

x^2-2xy-y^2=0
(x-y)^2=0
x-y=0
(x-y)/(x+y)=0