Đáp án:
$\begin{array}{l}
Dkxd:x > 0;x \ne 1\\
A = \dfrac{{\sqrt x + 2}}{{\sqrt x }}\\
\Rightarrow A + \sqrt x \\
= \dfrac{{\sqrt x + 2}}{{\sqrt x }} + \sqrt x \\
= 1 + \dfrac{2}{{\sqrt x }} + \sqrt x \\
Theo\,\text{Cô} - si:\\
\sqrt x + \dfrac{2}{{\sqrt x }} \ge 2.\sqrt {\sqrt x .\dfrac{2}{{\sqrt x }}} = 2\sqrt 2 \\
\Rightarrow \sqrt x + \dfrac{2}{{\sqrt x }} + 1 \ge 2\sqrt 2 + 1\\
\Rightarrow A + \sqrt x \ge 2\sqrt 2 + 1\\
\Rightarrow GTNN:A + \sqrt x = 2\sqrt 2 + 1\\
\text{Dấu = xảy ra khi:}:\sqrt x = \dfrac{2}{{\sqrt x }} \Rightarrow x = 2\left( {tmdk} \right)
\end{array}$
