数学代数题：(x/y+z)+(y/x+z)+(z/x+y)=1求(x²/y+z)+(y²/x+z)+(z²/x+y)=?

Question

数学代数题：(x/y+z)+(y/x+z)+(z/x+y)=1 求(x²/y+z)+(y²/x+z)+(z²/x+y)=?

Answer 1

x/(y+z)=1-[y/(z+x)+z/(x+y)]
y/(z+x)=1-[x/(y+z)+z/(x+y)]
z/(x+y)=1-[x/(y+z)+y/(z+x)]
x2/(y+z)+y2/(z+x)+z2/(x+y)
=x*[x/(y+z)]+y*[y/(z+x)]+z*[z/(x+y)]
=x*{1-[y/(z+x)+z/(x+y)]}+y*{1-[x/(y+z)+z/(x+y)]}+z*{1-[x/ (y+z)+y/(z+x)]}
=x-x*[y/(z+x)+z/(x+y)]+y-y*[x/(y+z)+z/(x+y)]+z-z*[x/(y+z)+y/(z+x)]
=x+y+z-[xy/(z+x)+xz/(x+y)+yx/(y+z)+yz/(x+y)+zx/(y+z)+zy/(z+x)]
=x+y+z-[xy/(z+x)+zy/(z+x)+yx/(y+z)+zx/(y+z)+xz/(x+y)+yz/(x+y)]
=x+y+z-[y(x+z)/(z+x)+x(y+z)/(y+z)+z(x+y)/(x+y)]
=x+y+z-(y+x+z)
=0