Để tìm max A, ta tìm min 1/A
\(\dfrac{1}{A}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}+1=2+1=3\)
\(\Rightarrow A\le\dfrac{1}{3}\)
Đẳng thức xảy ra \(\Leftrightarrow\sqrt{x}=\dfrac{1}{\sqrt{x}}\Leftrightarrow x=1\)