Question

设x,y,z都是实数,且x/y+z+y/z+x+z/x+y=0,求x^2/y+z+y^2/z+x+z^2/x+y

Answer 1

x+y+z=-(x/y+y/z+z/x)
原式=z+x+y+x²/y+y²/z+z²/x
=x²/y+y²/z+z²/x-(x/y+y/z+z/x)

x/y+y/z+z/x=(xy²+yz²+zx²)/xyz
x²/y+y²/z+z²/x=(x³z+y³x+z³y)/xyz

=(xy³+yz³+zx³)-(xy²+yz²+zx²)

(x+y+z)/(x/y+y/z+z/x)
=xyz(x+y+z)/(xy²+yz²+zx²)=-1

没有头绪了，标注此题，以后再解。

Answer 2

因为x/y+z + y/z+x +z/x+y=1
所以x/y+z=1-y/z+x-z/x+y，两边同乘以x
得x^2/y+z=x-xy/z+x-xz/x+y
同理y^2/x+z=y-xy/z+y-yz/x+y
,z^2=z-xz/y+z-yz/x+z
原式=x+y+z-(xy+zy)/x+z-(xz+yz)/x+y-(yx+zx)/y+z
=x+y+z-y-z-x
=0