Question

已知实数x,y,z满足x/(y+z)+y/(z+x)+z/(x+y)=1,则x平方/(y+z)+y平方/(z+x)+z平方/(x+y)=?

Answer 1

等于0.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+x-[y(x+z)/(z+x)+x(y+z)/(y+z)+z(x+y)/(x+y)] =x+y+z-(y+x+z) =0