Đáp án:
$\begin{array}{l}
B = \dfrac{{15\sqrt x - 11}}{{x + 2\sqrt x - 3}} - \dfrac{{2 - 3\sqrt x }}{{1 - \sqrt x }} - \dfrac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\
= \dfrac{{15}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}} + \dfrac{{2 - 3\sqrt x }}{{\sqrt x - 1}} - \dfrac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\
= \dfrac{{15 + \left( {2 - 3\sqrt x } \right)\left( {\sqrt x + 3} \right) - \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{15 + 2\sqrt x + 6 - 3x - 9\sqrt x - 2x + 2\sqrt x - 3\sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - 5x - 8\sqrt x + 24}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
A = \dfrac{{\left( {x + 2} \right){{\left( {x - 2} \right)}^2} - 9\left( {x + 2} \right)}}{{{{\left( {x - 5} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{\left( {x + 2} \right)\left( {{x^2} - 4x + 4 - 9} \right)}}{{{{\left( {x - 5} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{\left( {x + 2} \right)\left( {{x^2} - 4x - 5} \right)}}{{{{\left( {x - 5} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{\left( {x + 2} \right)\left( {x - 5} \right)\left( {x + 1} \right)}}{{{{\left( {x - 5} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{x + 2}}{{x - 5}}
\end{array}$
