题目

已知p＋q＋r＝9,且p/(x^2-yz)=q/(y^2-zx)=r/(z^2-xy) ,则(px+qy+rz)/(x+y+z)等于?
为什么能列出p/(x^2-yz)=q/(y^2-zx)=r/(z^2-xy)=(px+qy+rz)/(x^3+y^3+z^3-3xyz)?过程具体点.

提问时间：2020-12-30

答案

p/(x^2-yz)=px/(x^3-xyz)分子、分母同乘x q/(y^2-zx)=qy/(y^3-yzx)分子、分母同乘y r/(z^2-xy)=rz/(z^3-zxy)分子、分母同乘z (a/b=c/d=(a+c)/(b+d)比例性质) 所以 p/(x^2-yz)=q/(y^2-zx)=r/(z^2-xy)=(px+qy+rz)/(x^3+y^3+z^3-3xyz) p/(x^2-yz)=(px+qy+rz)/(x^3+y^3+z^3-3xyz) q/(y^2-zx)=(px+qy+rz)/(x^3+y^3+z^3-3xyz) r/(z^2-xy)=(px+qy+rz)/(x^3+y^3+z^3-3xyz) p=(x^2-yz)(px+qy+rz)/(x^3+y^3+z^3-3xyz) q=(y^2-zx)(px+qy+rz)/(x^3+y^3+z^3-3xyz) r=(z^2-xy)(px+qy+rz)/(x^3+y^3+z^3-3xyz) p+q+r=(px+qy+rz)(x^2+y^2+z^2-xy-xz-yz)/x^3+y^3+z^3-3xyz=9 (x+y+z)(x^2+y^2+z^2-xy-xz-yz)=x^3+y^3+z^3-3xyz (px+qy+rz)/(x+y+z)=9