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