a=2008,b=2007.c=2006,则多项式a^2+b^2+c^2-ab-ac-bcthanks!
问题描述：
a=2008,b=2007.c=2006,则多项式a^2+b^2+c^2-ab-ac-bc
thanks!
答
a=c+2 b=c+1
所以多项式=（c+2）²+(c+1)²+c²-(c+2)(c+1)-(c+2)c-(c+1)c=3(含c的项都消掉了)
答
a^2+b^2+c^2-ab-ac-bc
=1/2[(a-b)^2+(b-c)^2+(a-c)^2]
=1/2(1+1+4)
=3