Question

\(\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}< hoặc=0\)

Vậy X bằng.....

Answer 1

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 \(\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\le0\)

\(\Rightarrow\orbr{\begin{cases}\left|x-3\right|^{2014}=0\\\left|6+2y\right|^{2015}=0\end{cases}\Rightarrow\orbr{\begin{cases}x-3=0\\6+2y=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\y=-3\end{cases}}}\)