已知X-Y=2,Y-Z=1,试求X2+Y2+Z2-XY-YZ-ZX已知X-Y=2,Y-Z=1,试求X2+Y2+Z2-XY-YZ-ZX注：后面的X2,Y2,Z2的2是平方!

x-y=2 ①
y-z=1 ②
①+②得出 ： x-z=3 ③
①^2+②^2+③^2=(x-y)^2+(y-z)^2+(x-z)^2=14
=2x^2+2y^2+2z^2-2xy-2yz-2xz=14
所以x^2+y^2+z^2-xy-yz-xz=7

X-Y=2 X^2-2XY+Y^2=4 (1)
Y-Z=1 Y^2-2YZ+Z^2=1 (2)
X-Z=3 X^2-2XZ+Z^2=9 (3)
2X^2+2Y^2+2Z^2-2XY-2YZ-2XZ=14
X2+Y2+Z2-XY-YZ-ZX=7

由x-y=2,y-z=1
两式相加,得
x-z=3
所以
x^2+y^2+z^2-xy-yz-zx
=(2x^2+2y^2+2z^2-2xy-2yz-2zx)/2
=[(x^2+y^2-2xy)+(x^2+z^2-2zx)+(y^2+z^2-2zy)]/2
=[(x-y)^2+(x-z)^2+(y-z)^2]/2
=[2^2+1^2+3^2]/2
=7

x^2+y^2+z^2-xy-yz-zx
=(2x^2+2y^2+2z^2-2xy-2yz-2zx)/2
=[(x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)]/2
=[(x-y)^2+(y-z)^2+(z-x)^2]/2
x-y=2
y-z=1
z-x=(y-1)-(2+y)=y-1-2-y=-3
原式=(4+1+9)/2
=7