Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x\# 1\\
B = \dfrac{{15\sqrt x - 11}}{{x + 2\sqrt x - 3}} + \dfrac{{3\sqrt x - 2}}{{1 - \sqrt x }} - \dfrac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\
= \dfrac{{15\sqrt x - 11 - \left( {3\sqrt x - 2} \right)\left( {\sqrt x + 3} \right) - \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{15\sqrt x - 11 - \left( {3x + 7\sqrt x - 6} \right) - \left( {2x + \sqrt x - 3} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - 5x + 7\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - \left( {5\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}}\\
g)B = \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}}\\
= \dfrac{{ - 5\sqrt x - 15 + 17}}{{\sqrt x + 3}}\\
= \dfrac{{ - 5\left( {\sqrt x + 3} \right) + 17}}{{\sqrt x + 3}}\\
= - 5 + \dfrac{{17}}{{\sqrt x + 3}}\\
B \in Z\\
\Leftrightarrow \sqrt x + 3 = 17\left( {do:\sqrt x + 3 \ge 3} \right)\\
\Leftrightarrow \sqrt x = 14\\
\Leftrightarrow x = 196\left( {tmdk} \right)\\
Vậy\,x = 196\\
h)B = - 5 + \dfrac{{17}}{{\sqrt x + 3}}\\
Do:\sqrt x + 3 \ge 3\\
\Leftrightarrow \dfrac{1}{{\sqrt x + 3}} \le \dfrac{1}{3}\\
\Leftrightarrow \dfrac{{17}}{{\sqrt x + 3}} \le \dfrac{{17}}{3}\\
\Leftrightarrow - 5 + \dfrac{{17}}{{\sqrt x + 3}} \le \dfrac{2}{3}\\
\Leftrightarrow B \le \dfrac{2}{3}\\
GTLN:B = \dfrac{2}{3}\,Khi:x = 0
\end{array}$
