Question

已知a=2005x+2004，b=2005x+2005，c=2005x+2006则多项式a2+b2+c2-ab-bc-ac的值为

Answer 1

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

Answer 2

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

= (1/2)[1 + 1 + 4]

= 3