Đáp án:
$\begin{array}{l}
e)\dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{3\sqrt 2 }} + \dfrac{1}{{\sqrt 3 }}.\sqrt {\dfrac{5}{{12}} - \dfrac{1}{{\sqrt 6 }}} \\
= \dfrac{{\sqrt 3 }}{3} + \dfrac{{\sqrt 2 }}{6} + \dfrac{1}{{\sqrt 3 }}.\sqrt {\dfrac{{5 - 2\sqrt 6 }}{{12}}} \\
= \dfrac{{\sqrt 3 }}{3} + \dfrac{{\sqrt 2 }}{6} + \dfrac{1}{{\sqrt 3 }}.\dfrac{{\sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} }}{{2\sqrt 3 }}\\
= \dfrac{{\sqrt 3 }}{3} + \dfrac{{\sqrt 2 }}{6} + \dfrac{{\sqrt 3 - \sqrt 2 }}{6}\\
= \dfrac{{\sqrt 3 }}{3} + \dfrac{{\sqrt 3 }}{6}\\
= \dfrac{{\sqrt 3 }}{2}\\
a)C = \dfrac{{{a^4} - 4{a^2} + 3}}{{{a^4} - 12{a^2} + 27}}\\
= \dfrac{{{a^4} - {a^2} - 3{a^2} + 3}}{{{a^4} - 3{a^2} - 9{a^2} + 27}}\\
= \dfrac{{\left( {{a^2} - 1} \right)\left( {{a^2} - 3} \right)}}{{\left( {{a^2} - 3} \right)\left( {{a^2} - 9} \right)}}\\
= \dfrac{{{a^2} - 1}}{{{a^2} - 9}}\\
a = \sqrt 3 - \sqrt 2 \\
\Rightarrow {a^2} = 3 - 2\sqrt 6 + 2 = 5 - 2\sqrt 6 \\
\Rightarrow C = \dfrac{{5 - 2\sqrt 6 - 1}}{{5 - 2\sqrt 6 - 9}}\\
= \dfrac{{4 - 2\sqrt 6 }}{{ - 4 - 2\sqrt 6 }}\\
= 5 - 2\sqrt 6
\end{array}$
