已知x⼀(y+z)+y⼀(x+z)+z⼀(x+y)=1求x2⼀(y+z)+y2⼀(x+z）+z2⼀(x+y)的值

x/(y+z)+y/(x+z)+z/(x+y)=1
设x/(y+z)=a,y/(x+z)=b,z/(x+y)=c
∴x=ay+az
y=bx+bz
z=cx+cy
三式相加：
x+y+z=(b+c)x+(a+c)y+(a+b)z
∴[1-(b+c)]x+[1-(a+c)]y+[1-(a+b)]z=0
∵a+b+c=1
∴ax+by+cz=0
即x²/(y+z)+y²/(x+z）+z²/(x+y)=0

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

等于2啊，2{x/(y+z)+y/(x+z)+z/(x+y)}=2