Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
A = \left( {\dfrac{{\sqrt x - 2}}{{x - 1}} - \dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}} \right):\left( {\dfrac{{\sqrt x - 3}}{{x - 1}} - \dfrac{1}{{\sqrt x + 1}}} \right)\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}\\
:\dfrac{{\sqrt x - 3 - \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x - \sqrt x - 2 - \left( {x + \sqrt x - 2} \right)}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{ - 2}}\\
= \dfrac{{x - \sqrt x - 2 - x - \sqrt x + 2}}{{\sqrt x + 1}}.\dfrac{1}{{ - 2}}\\
= \dfrac{{ - 2\sqrt x }}{{\sqrt x + 1}}.\dfrac{1}{{ - 2}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
B = A.\dfrac{{\sqrt x + 1}}{{\sqrt x }} + \dfrac{{x - \sqrt x + 13}}{{\sqrt x + 3}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 1}}.\dfrac{{\sqrt x + 1}}{{\sqrt x }} + \dfrac{{x - \sqrt x + 13}}{{\sqrt x + 3}}\\
= 1 + \dfrac{{x - \sqrt x + 13}}{{\sqrt x + 3}}\\
= \dfrac{{\sqrt x + 3 + x - \sqrt x + 13}}{{\sqrt x + 3}}\\
= \dfrac{{x + 16}}{{\sqrt x + 3}}\\
\Rightarrow x + 16 = B.\sqrt x + 3B\\
\Rightarrow x - B.\sqrt x + 16 - 3B = 0\\
\Rightarrow \Delta \ge 0\\
\Rightarrow {B^2} - 4\left( {16 - 3B} \right) \ge 0\\
\Rightarrow {B^2} + 12B - 64 \ge 0\\
\Rightarrow \left( {B - 4} \right)\left( {B + 16} \right) \ge 0\\
\Rightarrow B \ge 4\\
\Rightarrow GTNN:B = 4\\
Khi:\sqrt x = 2 \Rightarrow x = 4\left( {tmdk} \right)
\end{array}$
