Cho các số thực x, y, z thỏa mãn...

Cách 1: Đặt $u={{3}^{x}}={{5}^{y}}={{15}^{\dfrac{2020}{x+y}-z}}\Rightarrow \left\{ \begin{aligned}
& u={{3}^{x}} \\
& u={{5}^{y}} \\
& u={{15}^{\dfrac{2020}{x+y}-z}} \\
\end{aligned} \right.$
$\Rightarrow \left\{ \begin{aligned}
& x={{\log }_{3}}u \\
& y={{\log }_{5}}u \\
& \left( \dfrac{2020}{x+y}-z \right)={{\log }_{15}}u \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& x=\dfrac{1}{{{\log }_{u}}3} \\
& y=\dfrac{1}{{{\log }_{u}}5} \\
& z=\dfrac{2020}{x+y}-{{\log }_{15}}u \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& x=\dfrac{1}{{{\log }_{u}}3} \\
& y=\dfrac{1}{{{\log }_{u}}5} \\
& z=\dfrac{2020}{{{\log }_{3}}u+{{\log }_{5}}u}-{{\log }_{15}}u \\
\end{aligned} \right.$
$\Rightarrow \left\{ \begin{aligned}
& xy=\dfrac{1}{{{\log }_{u}}3}.\dfrac{1}{{{\log }_{u}}5} \\
& yz=\dfrac{1}{{{\log }_{u}}5}\left( \dfrac{2020}{{{\log }_{3}}u+{{\log }_{5}}u}-{{\log }_{15}}u \right) \\
& zx=\dfrac{1}{{{\log }_{u}}3}\left( \dfrac{2020}{{{\log }_{3}}u+{{\log }_{5}}u}-{{\log }_{15}}u \right) \\
\end{aligned} \right.$
Khi đó: $S=xy+yz+zx=\dfrac{1}{{{\log }_{u}}3}.\dfrac{1}{{{\log }_{u}}5}+\dfrac{1}{{{\log }_{u}}5}\left( \dfrac{2020}{{{\log }_{3}}u+{{\log }_{5}}u}-{{\log }_{15}}u \right)$
$+\dfrac{1}{{{\log }_{u}}3}\left( \dfrac{2020}{{{\log }_{3}}u+{{\log }_{5}}u}-{{\log }_{15}}u \right)=2020$
Vậy $S=2020$
Cách 2: Chọn $x=1$. Do đó từ ${{3}^{x}}={{5}^{y}}={{15}^{\dfrac{2020}{x+y}-z}}\Rightarrow \left\{ \begin{aligned}
& y={{\log }_{5}}3 \\
& z=\dfrac{2020}{1+{{\log }_{5}}3}-{{\log }_{15}}3 \\
\end{aligned} \right.$
Do đó $S=xy+yz+zx$
$=1.{{\log }_{5}}3+{{\log }_{5}}3.\left( \dfrac{2020}{1+{{\log }_{5}}3}-{{\log }_{15}}3 \right)+\left( \dfrac{2020}{1+{{\log }_{5}}3}-{{\log }_{15}}3 \right).1=2020$
Lưu ý: Có thể dùng MTCT để tính S