Question

已知：a=2008x+2007,b=2008x+2008,c=2008x+2009,求a^2+b^2+c^2-ab-bc-ca的值

Answer 1

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。

Answer 2

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

Answer 3

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