已知x-y=4,y-z=2,求x^2+y^2+z^2-xy-yz-xz的值

x-y=4
y-z=2
两式相加：x-z=6
x^2+y^2+z^2-xy-yz-xz
=1/2*(x^2+y^2-2xy+x^2+z^2-2xz+y^2+z^2-2yz)
=1/2*[(x-y)^2+(x-z)^2+(y-z)^2]
=1/2*(4^2+6^2+2^2)
=1/2*(16+36+4)
=28