若x-3=y-2=z-1,求x²+y²+z²-xy-yz-xz的值

若x-3=y-2=z-1,求x²+y²+z²-xy-yz-xz的值

由已知条件可得：x-y=y-z=1,z-x=-2.且：2（x²+y²+z²-xy-yz-zx)=(x-y)²+(y-z)²+(z-x)²=1²+1²+(-2)²=6.所以：x²+y²+z²-xy-yz-zx=3.

x-3=y-2 x-y=1 y-2=z-1 y-z=1 x-3=z-1 z-x=-2 x^2+y^2+z^2-xy-yz-xz =x(x-y)+y(y-z)+z(z-x) =x+y-2z x-3=z-

由已知条件可得：x-y=y-z=1，z-x=-2。且：2（x²+y²+z²-xy-yz-zx)=(x-y)²+(y-z)²+(z-x)²=1²+1²+(-2)²=6。所以：x²+y²+z²-xy-yz-zx=3。