已知x-3=y-2=z-1,试求x²+y²+z²-xy-yz-zx的值

解
x-3=y-2
∴x-y=1
y-2=z-1
∴y-z=1
x-3=z-1
∴x-z=2
又2x²+2y²+2z²-2xy-2yz-2zx
=(x²-2xy+y²)+(x²-2zx+z²)+(y²-2yz+z²)
=(x-y)²+(x-z)²+(y-z)²
=1²+2²+1²
=6
∴原式=6/2=3