Ta có:
$\left\{\begin{array}{l}2^{3x}=5y^2-4y\\\dfrac{4^x+2^{x+1}}{2^x+2}=y\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}2^{3x}=5y^2-4y\\\dfrac{2^x(2^x+2)}{2^x+2}=y\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}2^x=y\\y^3=5y^2-4y\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}2^x=y\\y=1\end{array}\right.\\\left\{\begin{array}{l}2^x=y\\y=4\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=0\\y=1\end{array}\right.\\\left\{\begin{array}{l}x=2\\y=4\end{array}\right.\end{array}\right.$