已知x+3y-5z=0,x-3y+4z=0,且xyz≠0,求x²+y²+z²÷xy+yz+zx的值

题目

已知x+3y-5z=0,x-3y+4z=0,且xyz≠0,求x²+y²+z²÷xy+yz+zx的值

答案

x+3y-5z=0,
x-3y+4z=0
得到
x=z/2
y=3z/2
(x²+y²+z²)/(xy+yz+zx)
=(z²/4+9z²/4+z²)/(3z²/4+3z²/2+z²/2)
=14/11