三个式子相加，得6x+6y+6z=6,从而x+y+z=1
6x+6y+6z=6,从而x+y+z=1
z
等于1
解：x-3=y-2 x-y=1y-2=z-1 y-z=1x-z=2x²+y²+z²-xy-yz-zx=(1/2)(x²-2xy+y²+y²-2yz+z²+z²-2zx+x²)=(1/2)[(x-y)²+(y-z)²+(x-z)²]=(1/2)(1²+1²+2²)=3
