a²+b²+c²-ab-bc-ca
=(1/2)(a²-2ab+b²+b²-2bc+c²+c²-2ac+a²)
=(1/2)[(b-a)²+(c-b)²+(c-a)²]
=(1/2)[(2008x+2008-2008x-2007)²+(2008x+2009-2008x-2008)²+(2008x+2009-2008x-2007)²]
=(1/2)(1²+1²+2²)
=6/2
=3
为定值,与x的取值无关.