已知x+3y-5z=0,x-3y+4z=0,且xyz≠0,求x²+y²+z²÷xy+yz+zx的值
问题描述：
已知x+3y-5z=0,x-3y+4z=0,且xyz≠0,求x²+y²+z²÷xy+yz+zx的值
答
x+3y-5z=0 ----1
x-3y+4z=0 ----2
式2+1得
2x-z=0 -->x=z/2
式1-2得
6y-9z=0 -->y=3z/2
x²+y²+z²÷xy+yz+zx
=(z^2/4+9z^2/4+z^2)/(3z^2/4+3z^2/2+z^2/2) (将分式上下同乘以4得)
=(1+9+4)/(3+6+2)
=14/11
答
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