Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
Q = \left( {\dfrac{{15}}{{\sqrt 6 + 1}} + \dfrac{4}{{\sqrt 6 - 2}} - \dfrac{{12}}{{3 - \sqrt 6 }}} \right).\left( {\sqrt 6 + 1} \right)\\
= \left( {\dfrac{{15.\left( {\sqrt 6 - 1} \right)}}{{\left( {\sqrt 6 + 1} \right)\left( {\sqrt 6 - 1} \right)}} + \dfrac{{4.\left( {\sqrt 6 + 2} \right)}}{{\left( {\sqrt 6 - 2} \right)\left( {\sqrt 6 + 2} \right)}} - \dfrac{{12\left( {3 + \sqrt 6 } \right)}}{{\left( {3 - \sqrt 6 } \right)\left( {3 + \sqrt 6 } \right)}}} \right).\left( {\sqrt 6 + 1} \right)\\
= \left( {\dfrac{{15.\left( {\sqrt 6 - 1} \right)}}{{6 - 1}} + \dfrac{{4\left( {\sqrt 6 + 2} \right)}}{{6 - 4}} - \dfrac{{12.\left( {3 + \sqrt 6 } \right)}}{{9 - 6}}} \right).\left( {\sqrt 6 + 1} \right)\\
= \left[ {3.\left( {\sqrt 6 - 1} \right) + 2.\left( {\sqrt 6 + 2} \right) - 4.\left( {3 + \sqrt 6 } \right)} \right].\left( {\sqrt 6 + 1} \right)\\
= \left[ {3\sqrt 6 - 3 + 2\sqrt 6 + 4 - 12 - 4\sqrt 6 } \right].\left( {\sqrt 6 + 1} \right)\\
= \left( {\sqrt 6 - 11} \right)\left( {\sqrt 6 + 1} \right)\\
= 6 + \sqrt 6 - 11\sqrt 6 - 11\\
= - 5 - 10\sqrt 6
\end{array}\)
