已知a-b=2,a-c=1,求(2a-b-c)^2+(2b-a-c)^2+(2c-a-b)^2

a-b=2.......(1)
a-c=1.......(2)
(1) + (2)得： 2a-b-c=3
(2) - (1)×2得： 2b-a-c=-3
(1) - (2)×2得： 2c-a-b=0
(2a-b-c)^2+(2b-a-c)^2+(2c-a-b)^2 =3^2 + (-3)^2 + (0)^2 =18

a-b=2,a-c=1
c-b=1
(2a-b-c)^2+(2b-a-c)^2+(2c-a-b)^2
=((a-b)+(a-c))^2+((b-a+(b-c))^2+((c-a)+(c-b))^2
=(2+1)^2+(-2-1)^2+(-1+1)^2
=9+9+0
=18

2a-b-c)^2+(2b-a-c)^2+(2c-a-b)^2=(2+1)^2+(-4+1)^2+(-2+2)^2=18

如此即可
a-b=2,a-c=1
c-b=1
(2a-b-c)^2+(2b-a-c)^2+(2c-a-b)^2
=((a-b)+(a-c))^2+((b-a+(b-c))^2+((c-a)+(c-b))^2
=(2+1)^2+(-2-1)^2+(-1+1)^2
=9+9+0
=18