Question

已知a=2002,b=2003,c=2004,求a^2+b^2+c^2-ab-ac-bc的值

Answer 1

a^2+b^2+c^2-ab-bc-ac

=(2a^2+2b^2+2c^2-2ab-2bc-2ac)/2

=[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)]/2

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

=[(2002-2003)^2+(2003-2004)^2+(2004-2004)^2]/2

=(1+1+4)/2

=3