Vì \(\left| {3x + 5} \right| \ge 0\)
\(\left( {2y + 5} \right) \ge 0\)
\({\left( {4z - 3} \right)^{2020}} \ge 0\)
Do đó :
\(\begin{array}{l}\left| {3x + 5} \right| + {\left( {2y + 5} \right)^{2020}} + {\left( {4z - 3} \right)^{2020}} = 0\\ \Leftrightarrow \left\{ \begin{array}{l}3x + 5 = 0\\2y + 5 = 0\\4z - 3 = 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x = \frac{{ - 5}}{3}\\y = \frac{{ - 5}}{2}\\z = \frac{3}{4}\end{array} \right.\end{array}\)
Vậy \(x = \frac{{ - 5}}{3};\,y = \frac{{ - 5}}{2};z = \frac{3}{4}\)