初一数学问题，急用！！

a²+b²+c²-ab-bc-ac
=1/2(2a²+2b²+2c²-2ab-2bc-2ac)
=1/2[(a-b)²+(a-c)²+(b-c)²]
=3

a^2+b^2+c^2-ab-bc-ca
=a(a-b)+b(b-c)+c(c-a)
=a(1999x+2000-1999x-2001)+b(1999x+2001-1999x-2002)+c(1999x+2002-1999x-2000)
=-a-b+2c
=c-a+c-b
=1999x+2002-1999x-2000+1999x+2002-1999x-2001
=3

a²+b²+c²-ab-bc-ac
=[(a-b)²+(b-c)²+(c-a)²]/2
=[1+1+4]/2
=3

3

=2/1[(a-b)^2+(b-c)^2+(a-c)^2]
=3

因为2a²+2b²+2c²-2ab-2bc-2ac=（a-b）²+（b-c）²+（a-c）²
=1²+ 1²+2²=6
所以a²+b²+c²-ab-bc-ac=3