`(x - 2015)^(x + 2) - (x - 2015)^(x + 12) =0`

`=> (x - 2015)^(x + 2) - (x-2015)^10 . (x - 2015)^(x + 12) =0`

`=> (x - 2015)^(x + 2) [1 - (x - 2015)^10] =0`

`=>` $\left[\begin{matrix} (x - 2015)^{x + 2} = 0\\ 1- (x - 2015)^{10}= 0\end{matrix}\right.$

`=>` $\left[\begin{matrix} x -2015 = 0\\ (x - 2015)^{10}= 1\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=2015\\ x-2015=1\\x - 2015 =-1\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=2015\\ x=2016\\x  =2014\end{matrix}\right.$

Vậy `x∈{2014;2015;2016}`