Vì \(\left|x-3\right|^{2014}\ge0;\left|6+2y\right|^{2015}\ge0\)
\(\Rightarrow\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\ge0\)
Mà đề lại cho : \(\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\le0\Rightarrow\left|x-3\right|^{2014}=0;\left|6+2y\right|^{2015}=0\)
\(\Rightarrow x-3=0;6+2y=0\Rightarrow x=3;y=-3\)