Đáp án:
$\begin{array}{l}
c)P = \left( {\dfrac{{15}}{{\sqrt 6 + 1}} + \dfrac{4}{{\sqrt 6 - 2}} - \dfrac{{12}}{{3 - \sqrt 6 }}} \right).\left( {\sqrt 6 + 11} \right)\\
= \left( {\dfrac{{15\left( {\sqrt 6 - 1} \right)}}{{6 - 1}} + \dfrac{{4\left( {\sqrt 6 + 2} \right)}}{{6 - 4}} - \dfrac{{12\left( {3 + \sqrt 6 } \right)}}{{{3^2} - 6}}} \right).\left( {\sqrt 6 + 11} \right)\\
= \left( {3\left( {\sqrt 6 - 1} \right) + 2\left( {\sqrt 6 + 2} \right) - 4\left( {3 + \sqrt 6 } \right)} \right)\left( {\sqrt 6 + 11} \right)\\
= \left( {3\sqrt 6 - 3 + 2\sqrt 6 + 4 - 12 - 4\sqrt 6 } \right)\left( {\sqrt 6 + 11} \right)\\
= \left( {\sqrt 6 - 11} \right)\left( {\sqrt 6 + 11} \right)\\
= 6 - {11^2}\\
= 6 - 121\\
= - 115
\end{array}$
