Đáp án:
$\begin{array}{l}
a)\dfrac{{3\cot {{60}^0}}}{{2{{\cos }^2}{{30}^0} - 1}}\\
= \dfrac{{3.\dfrac{1}{{\sqrt 3 }}}}{{2.{{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2} - 1}} = \dfrac{{\sqrt 3 }}{{2.\dfrac{3}{4} - 1}} = 2\sqrt 3 \\
b)\dfrac{{\cos {{60}^0}}}{{1 + \sin {{60}^0}}} + \dfrac{1}{{\tan {{30}^0}}}\\
= \dfrac{{\dfrac{1}{2}}}{{1 + \dfrac{{\sqrt 3 }}{2}}} + \dfrac{1}{{\dfrac{1}{{\sqrt 3 }}}}\\
= \dfrac{{\dfrac{1}{2}}}{{\dfrac{{2 + \sqrt 3 }}{2}}} + \sqrt 3 \\
= \dfrac{1}{{2 + \sqrt 3 }} + \sqrt 3 \\
= \dfrac{{2 - \sqrt 3 }}{{{2^2} - 3}} + \sqrt 3 \\
= 2 - \sqrt 3 + \sqrt 3 \\
= 2
\end{array}$
