Biết rằng x, y là các số thực dương sao cho...

Ta có: ${{u}_{1}}={{2}^{3x}}{{\left( {{2}^{3}} \right)}^{{{\log }_{2}}y}}={{2}^{3x}}.{{y}^{3}}, {{u}_{2}}=\dfrac{{{2}^{x}}}{y}$
Theo bài ra ta có: $\left\{ \begin{aligned}
& {{2}^{3x}}.{{y}^{3}}.5y={{\left( \dfrac{{{2}^{x}}}{y} \right)}^{2}} \\
& {{2}^{3x}}.{{y}^{3}}+5y=2.\dfrac{{{2}^{x}}}{y} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{2}^{x}}.5{{y}^{6}}=1 \\
& {{2}^{3x}}.{{y}^{4}}+5{{y}^{2}}={{2.2}^{x}} \\
\end{aligned} \right.$.
Đặt $\left\{ \begin{aligned}
& u={{2}^{x}} \\
& v={{y}^{2}} \\
\end{aligned} \right.\left( u,v>0 \right)$ ta có:
$\left\{ \begin{aligned}
& 5u.{{v}^{3}}=1 \\
& {{u}^{3}}.{{v}^{2}}+5v=2u \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& u=\dfrac{1}{5{{v}^{3}}} \\
& \dfrac{1}{125{{v}^{7}}}+5v=\dfrac{2}{5{{v}^{3}}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& u=\dfrac{1}{5{{v}^{3}}} \\
& 625{{v}^{8}}-50{{v}^{4}}+1=0 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& v=\dfrac{1}{\sqrt{5}} \\
& u=\sqrt{5} \\
\end{aligned} \right.$.
Vậy ${{2}^{x}}.{{y}^{2}}=u.v=\dfrac{1}{\sqrt{5}}.\sqrt{5}=1$.