\(2009^{2011}+2011^{2009}=\left(2009^{2011}+1\right)+\left(2011^{2009}-1\right)\)
Ta có: \(a^n+b^n⋮\left(a+b\right)\) với n là số lẻ.
\(a^n-b^n⋮\left(a-b\right)\forall n\inℕ^∗\)
Nên \(2009^{2011}+1⋮\left(2009+1\right),2011^{2009}-1⋮\left(2011-1\right)\)
Vậy \(2009^{2011}+1+2011^{2009}-1⋮2010\Rightarrow2009^{2011}+2011^{2009}⋮2010\)