Ta có: \(\left|x+5\right|\ge0\forall x\)
\(\left(3y-4\right)^{2020}\ge0\forall y\)
Do đó: \(\left|x+5\right|+\left(3y-4\right)^{2020}\ge0\forall x,y\)
mà \(\left|x+5\right|+\left(3y-4\right)^{2020}=0\)
nên \(\left\{{}\begin{matrix}\left|x+5\right|=0\\\left(3y-4\right)^{2020}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\3y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=\frac{4}{3}\end{matrix}\right.\)
Vậy: x=-5 và \(y=\frac{4}{3}\)