ĐKXĐ; \(x\ge0\)
Dễ dàng nhận thấy \(A>0\)
\(A-3=\frac{\sqrt{x}+3}{2\sqrt{x}+1}-3=\frac{\sqrt{x}+3-6\sqrt{x}-3}{2\sqrt{x}+1}=\frac{-5\sqrt{x}}{2\sqrt{x}+1}\)
Do \(\left\{{}\begin{matrix}-5\sqrt{x}\le0\\2\sqrt{x}+1>0\end{matrix}\right.\) \(\Rightarrow A-3\le0\Rightarrow0< A\le3\)
Mà A nguyên \(\Rightarrow A=\left\{1;2;3\right\}\)
\(A=1\Rightarrow\frac{\sqrt{x}+3}{2\sqrt{x}+1}=1\Leftrightarrow\sqrt{x}+3=2\sqrt{x}+1\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
\(A=2\Rightarrow\sqrt{x}+3=4\sqrt{x}+2\Rightarrow3\sqrt{x}=1\Rightarrow x=\frac{1}{9}\)
\(A=3\Rightarrow\sqrt{x}+3=6\sqrt{x}+3\Rightarrow5\sqrt{x}=0\Rightarrow x=0\)
Vậy \(x=\left\{0;\frac{1}{9};4\right\}\)
