Lời giải:
Vì \(a=2018x+2015; b=2018x+2013; c=2019x+2019\)
\(\Rightarrow a-b=2; b-c=-x-6; c-a=x+4\)
Ta có:
\(a^2+b^2+c^2-ab-bc-ac=\frac{2a^2+2b^2+2c^2-2ab-2bc-2ac}{2}\)
\(=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}=\frac{2^2+(-x-6)^2+(x+4)^2}{2}\)
\(=\frac{2x^2+20x+56}{2}=x^2+10x+28\)