高分悬赏：问一道数学题（初三的）

展开得到X^2+Y^2+Z^2=XY+XZ+YZ
两边同时乘以2
2X^2+2Y^2+2Z^2=2XY+2XZ+2YZ
也就是(X-Y)^2+(X-Z)^2+(Y-Z)^2=0
所以X=Y=Z

3xy+3yz+3xz=(x+y+z)2
==>x^2+y^2+z^2=xy+xz+yz
(x-y)^2+(x-y)^2+(z-y)^2+==2(x^2+y^2+z^2)-2(xy+xz+yz)=0
==>x=y=z

展开
x^2+y^2+z^2+2xy+2yz+2zx=3xy+3yz+3zx
x^2+y^2+z^2-xyzx=0
2x^2+2y^2+2z^2-2xy-2xz-2yz=0
(x^2-2xy+y^2)+(x^2-2xz+z^2)+(y^2-2yz+z^2)=0
(x-y)^2+(x-z)^2+(y-z)^2=0
三式均不小于0
所以
x-y=0
y-z=0
x-z=0
所以x=y=z

由(3xy+3yz+3xz)*2=[(x+y+z)2]*2
可得(x-y)^2+(z-y)^2+(x-z)^2=0
所以 x-y=0
z-y=0
x-z=0

3xy+3yz+3xz=(x+y+z)2
3xy+3yz+3xz=x^2+y^2+z^2+2xy+2yz+2xz
x^2+y^2+z^2-xy-yz-xz=0
2x^2+2y^2+2z^2-2xy-2yz-2xz=0
(x^2+y^2-2xy)+(y^2+z^2-2yz)+(x^2+z^2-2xz)=0
(x-y)^2+(y-z)^2+(x-z)^2=0
所以x-y=0
y-z=0
x-z=0
所以x=y=z