TA CÓ:
\(\hept{\begin{cases}\left|x-2016\right|\ge0\Rightarrow\left|x-2016\right|^{2015}\ge0\\\left|y+2015\right|\ge0\Rightarrow\left|y+2015\right|^{2016}\ge0\end{cases}}.\)
Vậy\(\left|x+2016\right|^{2015}+\left|y+2016\right|^{2015}\ge0\)
be hon hoac bang ko ms dung ban a minh ghi nham