\(A>2A^2\\ \dfrac{\sqrt{x}+3}{\sqrt{x}}>2\left(\dfrac{\sqrt{x}+3}{\sqrt{x}}\right)^2\\ \dfrac{\sqrt{x}+3}{\sqrt{x}}>2.\dfrac{\left(\sqrt{x}+3\right)^2}{x}\\ \sqrt{x}\left(\sqrt{x}+3\right)>2\left(\sqrt{x}+3\right)^2\\ x+3\sqrt{x}>2\left(x+6\sqrt{x}+9\right)\\ x+3\sqrt{x}>2x+12\sqrt{x}+18\\ x+3\sqrt{x}-2x-12\sqrt{x}-18>0\\ -x-9\sqrt{x}-18>0\\ -x+3\sqrt{x}+6\sqrt{x}-18>0\\ \left(\sqrt{x}-3\right)\left(-\sqrt{x}+6\right)>0\\ TH1\left[{}\begin{matrix}\sqrt{x}-3>0\\-\sqrt{x}+6>0\end{matrix}\right.=>\left[{}\begin{matrix}x>9\\x< 36\end{matrix}\right.=>9< x< 36\)
\(TH2\\ \sqrt{x}-3< 0=>x< 9\\ -\sqrt{x}+6< 0\\ x>36\\ =>x< 9;x>36\left(voli\right)\)
Theo bài ra, ta có: \(A>2A^2\)
\(\Rightarrow\) \(\dfrac{\sqrt{x}+3}{\sqrt{x}}>\dfrac{2(\sqrt{x}+3)^2}{x}\)
\(\Rightarrow\) \(\dfrac{x+3\sqrt{x}-2(\sqrt{x}+3)^2}{x}>0\)
\(\Rightarrow\) \(\dfrac{x+3\sqrt{x}-2x-12\sqrt{x}-18}{x}>0\)
\(\Rightarrow\) \(\dfrac{-x-9\sqrt{x}-18}{x}>0\) mà \(x>0\)
\(\Rightarrow\) \(-x-9\sqrt{x}-18>0\)
\(\Rightarrow\) \(x+9\sqrt{x}+18<0\)
\(\Rightarrow\) \(-6<\sqrt{x}<-3\) (vô lí)
Do đó không tồn tại giá trị nào của \(x\) thỏa mãn \(A>2A^2\)
