Đáp án:
$\begin{array}{l}
A = \left( {\dfrac{1}{{2 + \sqrt 3 }} - \dfrac{{12}}{{3 + \sqrt 3 }} + \dfrac{{26}}{{4 - \sqrt 3 }}} \right)\left( {\sqrt 4 - 3\sqrt 3 } \right)\\
= \left( \begin{array}{l}
\dfrac{{2 - \sqrt 3 }}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}} - \dfrac{{12\left( {3 - \sqrt 3 } \right)}}{{\left( {3 + \sqrt 3 } \right)\left( {3 - \sqrt 3 } \right)}}\\
+ \dfrac{{26\left( {4 + \sqrt 3 } \right)}}{{\left( {4 + \sqrt 3 } \right)\left( {4 - \sqrt 3 } \right)}}
\end{array} \right)\\
.\left( {2 - 3\sqrt 3 } \right)\\
= \left( {\dfrac{{2 - \sqrt 3 }}{{4 - 3}} - \dfrac{{36 - 12\sqrt 3 }}{{9 - 3}} + \dfrac{{104 + 26\sqrt 3 }}{{16 - 3}}} \right).\left( {2 - 3\sqrt 3 } \right)\\
= \left( {2 - \sqrt 3 - 6 + 2\sqrt 3 + 8 + 2\sqrt 3 } \right).\left( {2 - 3\sqrt 3 } \right)\\
= \left( {4 + 3\sqrt 3 } \right)\left( {2 - 3\sqrt 3 } \right)\\
= 8 - 12\sqrt 3 + 6\sqrt 3 - 27\\
= - 19 - 6\sqrt 3
\end{array}$
