Question

a=2009x+2008、b=2009x+2009、c=2009x+2010、求a^2+b^2+c^2-ab-bc-ac

Answer 1

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]

代入得：原式=1/2[(-1)^2+(-1)^2+(-2)^2]

=1/2(1+1+4)=3