Question

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

Answer 1

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

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Answer 2

x+3y-5z=0 ----1
x-3y+4z=0 ----2
式2+1得
2x-z=0 -->x=z/2

式橘搭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) (将圆碰拿分式上吵渣下同乘以4得)
=(1+9+4)/(3+6+2)
=14/11