Điều kiện: $x,y\ne0$
Ta có:
$\left\{\begin{array}{l}\dfrac{2x}{y}+\dfrac{2y}{x}=5\\x^2+x-2y=4\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}2x^2-5xy+2y^2=0\\x^2+x-2y=4\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=2y\\x^2+x-2y=4\end{array}\right.\\\left\{\begin{array}{l}2x=y\\x^2+x-2y=4\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=2y\\x^2=4\end{array}\right.\\\left\{\begin{array}{l}2x=y\\x^2-3x-4=0\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=2\\y=1\end{array}\right.\\\left\{\begin{array}{l}x=-2\\y=-1\end{array}\right.\\\left\{\begin{array}{l}x=-1\\y=-2\end{array}\right.\\\left\{\begin{array}{l}x=4\\y=8\end{array}\right.\end{array}\right.$