Đáp án:
$\begin{array}{l}
a)A = \dfrac{{x + 16}}{{2\sqrt x }}\\
= \dfrac{{\sqrt x }}{2} + \dfrac{8}{{\sqrt x }}\\
Theo\,Co - si:\\
\dfrac{{\sqrt x }}{2} + \dfrac{8}{{\sqrt x }} \ge 2.\sqrt {\dfrac{{\sqrt x }}{2}.\dfrac{8}{{\sqrt x }}} = 2\sqrt 4 = 4\\
\Leftrightarrow A \ge 4\\
\Leftrightarrow GTNN:A = 4\,\\
Khi:\dfrac{{\sqrt x }}{2} = \dfrac{8}{{\sqrt x }} \Leftrightarrow x = 16\left( {tmdk} \right)\\
b)B = \dfrac{{x - 3\sqrt x + 4}}{{\sqrt x }}\\
= \sqrt x - 3 + \dfrac{4}{{\sqrt x }}\\
= \sqrt x + \dfrac{4}{{\sqrt x }} - 3\\
\ge 2.\sqrt {\sqrt x .\dfrac{4}{{\sqrt x }}} - 3\\
\Leftrightarrow B \ge 2\sqrt 4 - 3 = 1\\
\Leftrightarrow GTNN:B = 1\\
Khi:\sqrt x = \dfrac{4}{{\sqrt x }} \Leftrightarrow x = 4\left( {tmdk} \right)\\
c)C = \dfrac{{x + \sqrt x + 9}}{{\sqrt x }}\\
= \sqrt x + 1 + \dfrac{9}{{\sqrt x }}\\
= \sqrt x + \dfrac{9}{{\sqrt x }} + 1\\
\ge 2\sqrt {\sqrt x .\dfrac{9}{{\sqrt x }}} + 1\left( {theo\,Co - si} \right)\\
\Leftrightarrow C \ge 2\sqrt 9 + 1\\
\Leftrightarrow C \ge 7\\
\Leftrightarrow GTNN:C = 7\,khi:\sqrt x = \dfrac{9}{{\sqrt x }} \Leftrightarrow x = 9
\end{array}$
