Ta có:
$5\sqrt x+\dfrac{5}{2\sqrt x}=2x+\dfrac{1}{2x}+4$
$\Leftrightarrow 5\left(\sqrt x+\dfrac{1}{2\sqrt x}\right)=2\left(x+\dfrac{1}{4x}\right)+4$
$\Leftrightarrow 5\left(\sqrt x+\dfrac{1}{2\sqrt x}\right)=2\left(\sqrt x+\dfrac{1}{2\sqrt x}\right)^2+2$
$\Leftrightarrow \left[\begin{array}{l}\sqrt x+\dfrac{1}{2\sqrt x}=2\\\sqrt x+\dfrac{1}{2\sqrt x}=\dfrac{1}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{3}{2}+\sqrt2\\x=\dfrac{3}{2}-\sqrt2\end{array}\right.$