Đáp án:
$\begin{array}{l}
A = \frac{{x + \sqrt x + 4}}{{\sqrt x + 1}} = \frac{{\sqrt x \left( {\sqrt x + 1} \right) + 4}}{{\sqrt x + 1}} = \sqrt x + 1 + \frac{4}{{\sqrt x + 1}} - 1\\
Do:\sqrt x \ge 0\forall x\\
\Rightarrow \sqrt x + 1 > 0\forall x\\
\Rightarrow Theo\,Co - si:\\
\left( {\sqrt x + 1} \right) + \frac{4}{{\sqrt x + 1}} \ge 2\sqrt {\left( {\sqrt x + 1} \right).\frac{4}{{\sqrt x + 1}}} = 4\\
\Rightarrow A \ge 4 - 1 = 3\\
\Rightarrow GTNN:A = 3 \Leftrightarrow x = 0
\end{array}$
