Question

若a=2003 b=2004 c=2005,求a^+b^+c^-ab-bc-ac的值

Answer 1

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

Answer 2

a-b=-1
b-c=-1
c-a=2

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