Đáp án:
$\begin{array}{l}
b)\dfrac{4}{{\sqrt 3 - 1}} - \dfrac{8}{{\sqrt 3 + 1}} - \dfrac{{11}}{{2\sqrt 3 - 1}}\\
= \dfrac{{4\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} - \dfrac{{8\left( {\sqrt 3 - 1} \right)}}{{3 - 1}} - \dfrac{{11\left( {2\sqrt 3 + 1} \right)}}{{{{\left( {2\sqrt 3 } \right)}^2} - 1}}\\
= \dfrac{{4\left( {\sqrt 3 + 1} \right)}}{2} - \dfrac{{8\left( {\sqrt 3 - 1} \right)}}{2} - \dfrac{{11\left( {2\sqrt 3 + 1} \right)}}{{11}}\\
= 2\left( {\sqrt 3 + 1} \right) - 4\left( {\sqrt 3 - 1} \right) - \left( {2\sqrt 3 + 1} \right)\\
= 2\sqrt 3 + 2 - 4\sqrt 3 + 4 - 2\sqrt 3 - 1\\
= 5 - 4\sqrt 3 \\
c)\sqrt {9{a^6}} - 3{a^3}\left( {a < 0} \right)\\
= - 3{a^3} - 3{a^3}\left( {a < 0} \right)\\
= - 6{a^3}\\
d)A = \sqrt {2 - \sqrt 3 } - \sqrt {2 + \sqrt 3 } \\
\Leftrightarrow {A^2} = 2 - \sqrt 3 - 2\sqrt {2 - \sqrt 3 } .\sqrt {2 + \sqrt 3 } + 2 + \sqrt 3 \\
= 4 - 2\sqrt {{2^2} - 3} \\
= 4 - 2\\
= 2\\
\Leftrightarrow A = - \sqrt 2 \left( {do:A < 0} \right)\\
\Leftrightarrow \sqrt {2 - \sqrt 3 } - \sqrt {2 + \sqrt 3 } = - \sqrt 2
\end{array}$
