\(\left(24-4y\right)^{2018}+\left|x^2-4\right|^{2019}\le0\left(1\right)\)
Vì \(\hept{\begin{cases}\left(24-4y\right)^{2018}\ge0;\forall x,y\\\left|x^2-4\right|^{2019}\ge0;\forall x,y\end{cases}}\)\(\Rightarrow\left(24-4y\right)^{2018}+\left|x^2-4\right|^{2019}\ge0;\forall x,y\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}\left(24-4y\right)^{2018}=0\\\left|x^2-4\right|^{2019}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y=6\\x=\pm2\end{cases}}\)
Vậy \(\left(x,y\right)\in\left\{\left(2;6\right);\left(-2;6\right)\right\}\)
