\(\left|2x-28\right|^{2018}+\left(3y+9\right)^{2020}=0\)
Giải:Vì \(\left|2x-28\right|^{2018}\ge0\);\(\left(3y+9\right)^{2020}\ge0\)
\(\Rightarrow\left|2x-28\right|^{2018}+\left(3y+9\right)^{2020}\ge0\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left|2x-28\right|=0\\3y+9=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x-28=0\\3y+9=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=14\\y=-3\end{cases}}\)