Ta có:
$\begin{cases}\left(3x -\dfrac{5}{9}\right)^{2018} \geq 0\,\,\forall x\\\left(3y +\dfrac{0,4}{3}\right)^{2020} \geq 0\,\,\forall y\end{cases}$
Do đó:
$\left(3x -\dfrac{5}{9}\right)^{2018} +\left(3y +\dfrac{0,4}{3}\right)^{2020} =0$
$\Leftrightarrow \begin{cases}3x - \dfrac59 = 0\\3y + \dfrac{0,4}{3} =0\end{cases}$
$\Leftrightarrow \begin{cases}x =\dfrac{5}{9}:3 =\dfrac{5}{27}\\y = -\dfrac{0,4}{3}:3 = -\dfrac{2}{45}\end{cases}$