\(\frac{2\sqrt{x}-1}{\sqrt{x}+2}=\frac{2\sqrt{x}+4-5}{\sqrt{x}+2}=\frac{2\left(\sqrt{x}+2\right)-5}{\sqrt{x}+2}=2-\frac{5}{\sqrt{x}+2}\)
Để
\(\Rightarrow\frac{5}{\sqrt{x}+2}\in Z\)
\(\Rightarrow5⋮\sqrt{x}+2\)
\(\Rightarrow\sqrt{x}+2\in\left(-1;1;-5;5\right)\)
\(\Rightarrow\sqrt{x}\in\left(-3;-1;-7;3\right)\)
\(\Rightarrow x\in\left(9;1;49\right)\)