Ta có:\(\hept{\begin{cases}\left|x-3\right|\ge0\\\left|6+2y\right|\ge0\end{cases}\Rightarrow\hept{\begin{cases}\left|x-3\right|^{2014}\ge0\\\left|6+2y\right|^{2015}\ge0\end{cases}\Rightarrow}\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\ge0}\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left|x-3\right|^{2014}=0\\\left|6-2y\right|^{2015}=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=3\end{cases}}}\)