Đáp án:
$\begin{array}{l}
Dkxd:x > 4;x \ne 9\\
B = \dfrac{{ - 3}}{{\sqrt x - 2}}\left( {x + 1} \right)\\
= \dfrac{{ - 3\left( {x + 1} \right)}}{{\sqrt x - 2}}\\
= \dfrac{{ - 3x - 3}}{{\sqrt x - 2}}\\
= \dfrac{{ - 3x + 12 - 12 - 3}}{{\sqrt x - 2}}\\
= \dfrac{{ - 3\left( {x - 4} \right) - 15}}{{\sqrt x - 2}}\\
= \dfrac{{ - 3\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right) - 15}}{{\sqrt x - 2}}\\
= - 3\left( {\sqrt x + 2} \right) - \dfrac{{15}}{{\sqrt x - 2}}\\
= - 3\left( {\sqrt x - 2} \right) - \dfrac{{15}}{{\sqrt x - 2}} - 6 - 6\\
= - \left[ {3\left( {\sqrt x - 2} \right) + \dfrac{{15}}{{\sqrt x - 2}}} \right] - 12\\
Do:x > 4 \Rightarrow \sqrt x - 2 > 0\\
Theo\,Co - si:\\
3\left( {\sqrt x - 2} \right) + \dfrac{{15}}{{\sqrt x - 2}} \ge 2.\sqrt {3\left( {\sqrt x - 2} \right).\dfrac{{15}}{{\sqrt x - 2}}} \\
\Rightarrow 3\left( {\sqrt x - 2} \right) + \dfrac{{15}}{{\sqrt x - 2}} \ge 2.3.\sqrt 5 = 6\sqrt 5 \\
\Rightarrow - \left[ {3\left( {\sqrt x - 2} \right) + \dfrac{{15}}{{\sqrt x - 2}}} \right] \le - 6\sqrt 5 \\
\Rightarrow B \le - 6\sqrt 5 - 12\\
\Rightarrow GTLN:B = - 6\sqrt 5 - 12\\
Khi:3\left( {\sqrt x - 2} \right) = \dfrac{{15}}{{\sqrt x - 2}}\\
\Rightarrow {\left( {\sqrt x - 2} \right)^2} = 5\\
\Rightarrow \sqrt x - 2 = \sqrt 5 \\
\Rightarrow \sqrt x = 2 + \sqrt 5 \\
\Rightarrow x = 9 + 4\sqrt 5 \left( {tmdk} \right)
\end{array}$
