a=1999x+2000
b=1999x+2001
c=1999x+2002
所以,a-b=-1;b-c=-1,;a-c=-2
a^2+b^2+c^2-ab-bc-ac
=(1/2)*(2a^2+2b^2+2c^2-2ab-2bc-2ac)
=(1/2)*[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)]
=(1/2)*[(a-b)^2+(b-c)^2+(a-c)^2]
=(1/2)*(1+1+4)
=3
a=1999x+2000
b=1999x+2001
c=1999x+2002
所以,a-b=-1;b-c=-1,;a-c=-2
a^2+b^2+c^2-ab-bc-ac
=(1/2)*(2a^2+2b^2+2c^2-2ab-2bc-2ac)
=(1/2)*[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)]
=(1/2)*[(a-b)^2+(b-c)^2+(a-c)^2]
=(1/2)*(1+1+4)
=3