已知x^2+根号2*y=根号3,y^2+根号2*x=根号3.且x不等于y,求y/x+x/y的值

两式想减
x^2-y^2+√2y-√x=0
(x+y)(x-y)-√2(x-y)=0
x不等于y则x-y不等于0
把x-y约分
x+y-√2=0
x+y=√2

两式相加
(x^2+y^2)+√2(x+y)=2√3
(x+y)^2-2xy+√2(x+y)=2√3
x+y=√2
所以2-2xy+2=2√3
xy=2-√3

y/x+x/y
=(x^2+y^2)/xy
=[(x+y)^2-2xy]/xy
=(2-4+2√3)/(2-√3)
=2√3+2

x^2+根号2*y=根号3,y^2+根号2*x=根号3.
x^2+√2y=y^2+√2x
(x^2-y^2)-√2(x-y)=0
(x-y)(x+y-√2)=0
因为x不等于y，所以，x+y=√2
(x+y)^2=2
x^2+y^2+2xy=2

而：x^2+y^2+√2(x+y)=2√3
x^2+y^2+√2*√2=2√3
x^2+y^2=2√3-2

所以，(2√3-2)+2xy=2
xy=2-√3

y/x+x/y
=(x^2+y^2)/xy
=(2√3-2)/(2-√3)
=(2√3-2)(2+√3)
=2+2√3