Đáp án:
$\begin{array}{l}
Dkxd:x > 0;x \ne 9\\
1)x = 1,44\\
\Leftrightarrow \sqrt x = 1,2\\
A = \dfrac{{x + 7}}{{\sqrt x }} = \dfrac{{1,44 + 7}}{{1,2}} = \dfrac{{8,44}}{{1,2}} = \dfrac{{211}}{{30}}\\
2)B = \dfrac{{\sqrt x }}{{\sqrt x + 3}} + \dfrac{{2\sqrt x - 1}}{{\sqrt x - 3}} - \dfrac{{2x - \sqrt x - 3}}{{x - 9}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 3} \right) + \left( {2\sqrt x - 1} \right)\left( {\sqrt x + 3} \right) - 2x + \sqrt x + 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x - 3\sqrt x + 2x + 6\sqrt x - \sqrt x - 3 - 2x + \sqrt x + 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x + 3\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 3}}\\
3)S = \dfrac{1}{B} + A\\
= \dfrac{{\sqrt x - 3}}{{\sqrt x }} + \dfrac{{x + 7}}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x + 4}}{{\sqrt x }}\\
= \sqrt x + 1 + \dfrac{4}{{\sqrt x }}\\
Theo\,Co - si:\\
\sqrt x + \dfrac{4}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{4}{{\sqrt x }}} = 4\\
\Leftrightarrow \sqrt x + \dfrac{4}{{\sqrt x }} + 1 \ge 5\\
\Leftrightarrow S \ge 5\\
\Leftrightarrow GTNN:S = 5\,khi:x = 4\left( {tmdk} \right)
\end{array}$
