\(A=\dfrac{2\sqrt{n}-1}{\sqrt{n}-4}=\dfrac{2\sqrt{n}-8+7}{\sqrt{n}-4}=2+\dfrac{7}{\sqrt{n}+4}\)
Để A nguyên thì \(7⋮\sqrt{n}-4\)
\(\sqrt{n}-4\in\left\{-7;-1;1;7\right\}\\\Leftrightarrow\sqrt{n}\in\left\{-3;2;4;11\right\}\\ \Leftrightarrow n\in\left\{9;4;16;121\right\}\)