Đáp án:
$\begin{array}{l}
c)Dkxd:x > 0;x \ne 9\\
A = \dfrac{{x + 7}}{{\sqrt x }};B = \dfrac{{\sqrt x }}{{\sqrt x - 3}}\\
S = \dfrac{1}{B} + A\\
= 1:\dfrac{{\sqrt x }}{{\sqrt x - 3}} + \dfrac{{x + 7}}{{\sqrt x }}\\
= \dfrac{{\sqrt x - 3}}{{\sqrt x }} + \dfrac{{x + 7}}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x + 4}}{{\sqrt x }}\\
= \sqrt x + 1 + \dfrac{4}{{\sqrt x }}\\
Do:\sqrt x > 0\\
Theo\,Co - si:\\
\sqrt x + \dfrac{4}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{4}{{\sqrt x }}} = 4\\
\Leftrightarrow \sqrt x + \dfrac{4}{{\sqrt x }} + 1 \ge 5\\
\Leftrightarrow S \ge 5\\
\Leftrightarrow GTNN:S = 5\,khi:x = 4
\end{array}$
