已知x,y,z满足x/(y+z)+y/(z+x)+z/(x+y)=1,求代数式x2/(y+z)+y2/(x+z)+z2/

1个回答

  • x/(y+z)+y/(z+x)+z/(x+y)=1

    所以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+zy)/(z+x)+(yx+zx)/(y+z)+(xz+yz)/(x+y)]

    =x+y+x-[y(x+z)/(z+x)+x(y+z)/(y+z)+z(x+y)/(x+y)]

    =x+y+z-(y+x+z)

    =0