Đáp án:
$\begin{array}{l}
Dkxd:x > - 1\\
B = \left( {\dfrac{{\sqrt {x + 1} + 4}}{{\sqrt {x + 1} - 1}} - \dfrac{{2\sqrt {x + 1} - 4}}{{\sqrt {x + 1} + 1}}} \right).x\\
= \dfrac{{\left( {\sqrt {x + 1} + 4} \right)\left( {\sqrt {x + 1} + 1} \right) - \left( {2\sqrt {x + 1} - 4} \right)\left( {\sqrt {x + 1} - 1} \right)}}{{\left( {\sqrt {x + 1} - 1} \right)\left( {\sqrt {x + 1} + 1} \right)}}.x\\
= \dfrac{{x + 1 + 5\sqrt {x + 1} + 4 - \left( {2x + 2 - 6\sqrt {x + 1} + 4} \right)}}{{x + 1 - 1}}.x\\
= \dfrac{{x + 5\sqrt {x + 1} + 5 - 2x + 6\sqrt {x + 1} - 6}}{x}.x\\
= - x + 11\sqrt {x + 1} - 1\\
Khi:x = 4\left( {tmdk} \right)\\
\Leftrightarrow B = - 4 + 11\sqrt {4 + 1} - 1 = 11\sqrt 5 - 5\\
C = \left( {\dfrac{{\sqrt {x + 3} - 2}}{{\sqrt {x + 3} - 1}} - \dfrac{{2\sqrt {x + 3} - 5}}{{\sqrt {x + 3} + 1}}} \right)\left( {x - 2} \right)\\
= \dfrac{{\left( {\sqrt {x + 3} - 2} \right)\left( {\sqrt {x + 3} + 1} \right) - \left( {2\sqrt {x + 3} - 5} \right)\left( {\sqrt {x + 3} - 1} \right)}}{{\left( {\sqrt {x + 3} + 1} \right)\left( {\sqrt {x + 3} - 1} \right)}}\\
.\left( {x - 2} \right)\\
= \dfrac{{x + 3 - \sqrt {x + 3} - 2 - \left( {2x + 6 - 7\sqrt {x + 3} + 5} \right)}}{{x + 3 - 1}}.\left( {x - 2} \right)\\
= \dfrac{{x + 1 - \sqrt {x + 3} - 2x + 7\sqrt {x + 3} - 11}}{{x - 2}}.\left( {x - 2} \right)\\
= - x + 6\sqrt {x + 3} - 10\\
Khi:x = 4\\
\Leftrightarrow C = - 4 + 6\sqrt {4 + 3} - 10\\
= 6\sqrt 7 - 14
\end{array}$
