Đáp án:
$\begin{array}{l}
1)\\
a)x = 16\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = 4\\
\Leftrightarrow B = \dfrac{1}{{\sqrt x - 3}} = \dfrac{1}{{4 - 3}} = 1\\
b)M = A:B\\
= \left( {\dfrac{{x + \sqrt x + 10}}{{x - 9}} - \dfrac{1}{{\sqrt x - 3}}} \right):\dfrac{1}{{\sqrt x - 3}}\\
= \dfrac{{x + \sqrt x + 10 - \sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\left( {\sqrt x - 3} \right)\\
= \dfrac{{x + 7}}{{\sqrt x + 3}}\\
c)M = \dfrac{{x + 7}}{{\sqrt x + 3}} = \dfrac{{x - 9 + 16}}{{\sqrt x + 3}}\\
= \sqrt x - 3 + \dfrac{{16}}{{\sqrt x + 3}}\\
= \sqrt x + 3 + \dfrac{{16}}{{\sqrt x + 3}} - 6\\
Theo\,Co - si:\\
\sqrt x + 3 + \dfrac{{16}}{{\sqrt x + 3}} \ge 2\sqrt {\left( {\sqrt x + 3} \right).\dfrac{{16}}{{\sqrt x + 3}}} = 8\\
\Leftrightarrow \sqrt x + 3 + \dfrac{{16}}{{\sqrt x + 3}} - 6 \ge 2\\
\Leftrightarrow M \ge 2\\
\Leftrightarrow GTNN:M = 2\,\\
Khi:\sqrt x + 2 = 4 \Leftrightarrow x = 4\left( {tm} \right)\\
2)a)x = 9\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = 3\\
A = \dfrac{{1 - \sqrt x }}{{1 + \sqrt x }} = \dfrac{{1 - 3}}{{1 + 3}} = \dfrac{{ - 1}}{2}\\
b)B = \left( {\dfrac{{6 - \sqrt x }}{{x - 4}} + \dfrac{2}{{\sqrt x + 2}}} \right):\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}}\\
= \dfrac{{6 - \sqrt x + 2\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\
= \dfrac{{6 - \sqrt x + 2\sqrt x - 4}}{{\sqrt x + 2}}.\dfrac{1}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x + 2}}.\dfrac{1}{{\sqrt x + 1}}\\
= \dfrac{1}{{\sqrt x + 1}}\\
c)B - A \le \dfrac{3}{5}\\
\Leftrightarrow \dfrac{1}{{\sqrt x + 1}} - \dfrac{{1 - \sqrt x }}{{1 + \sqrt x }} \le \dfrac{3}{5}\\
\Leftrightarrow \dfrac{{\sqrt x }}{{\sqrt x + 1}} \le \dfrac{3}{5}\\
\Leftrightarrow 5\sqrt x \le 3\sqrt x + 3\\
\Leftrightarrow 2\sqrt x \le 3\\
\Leftrightarrow \sqrt x \le \dfrac{3}{2}\\
\Leftrightarrow x \le \dfrac{9}{4}\\
Vậy\,0 \le x \le \dfrac{9}{4}
\end{array}$
