Đáp án:

\(\begin{array}{l}
a,\,\,\,\, - \sqrt 3 \\
b,\,\,\,\,2\sqrt 6 \\
c,\,\,\,\,14\sqrt 3  - 24
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
\dfrac{1}{{1 - \sqrt 3 }} - \dfrac{1}{{1 + \sqrt 3 }}\\
 = \dfrac{{1.\left( {1 + \sqrt 3 } \right) - 1.\left( {1 - \sqrt 3 } \right)}}{{\left( {1 - \sqrt 3 } \right)\left( {1 + \sqrt 3 } \right)}}\\
 = \dfrac{{1 + \sqrt 3  - 1 + \sqrt 3 }}{{{1^2} - {{\sqrt 3 }^2}}}\\
 = \dfrac{{2\sqrt 3 }}{{1 - 3}}\\
 = \dfrac{{2\sqrt 3 }}{{ - 2}}\\
 =  - \sqrt 3 \\
b,\\
\dfrac{3}{{\sqrt 5  - \sqrt 2 }} + \dfrac{4}{{\sqrt 6  + \sqrt 2 }} + \dfrac{1}{{\sqrt 6  + \sqrt 5 }}\\
 = \dfrac{{3\left( {\sqrt 5  + \sqrt 2 } \right)}}{{\left( {\sqrt 5  - \sqrt 2 } \right)\left( {\sqrt 5  + \sqrt 2 } \right)}} + \dfrac{{4.\left( {\sqrt 6  - \sqrt 2 } \right)}}{{\left( {\sqrt 6  + \sqrt 2 } \right)\left( {\sqrt 6  - \sqrt 2 } \right)}} + \dfrac{{1.\left( {\sqrt 6  - \sqrt 5 } \right)}}{{\left( {\sqrt 6  + \sqrt 5 } \right)\left( {\sqrt 6  - \sqrt 5 } \right)}}\\
 = \dfrac{{3\left( {\sqrt 5  + \sqrt 2 } \right)}}{{{{\sqrt 5 }^2} - {{\sqrt 2 }^2}}} + \dfrac{{4.\left( {\sqrt 6  - \sqrt 2 } \right)}}{{{{\sqrt 6 }^2} - {{\sqrt 2 }^2}}} + \dfrac{{\sqrt 6  - \sqrt 5 }}{{{{\sqrt 6 }^2} - {{\sqrt 5 }^2}}}\\
 = \dfrac{{3\left( {\sqrt 5  + \sqrt 2 } \right)}}{{5 - 2}} + \dfrac{{4.\left( {\sqrt 6  - \sqrt 2 } \right)}}{{6 - 2}} + \dfrac{{\sqrt 6  - \sqrt 5 }}{{6 - 5}}\\
 = \dfrac{{3\left( {\sqrt 5  + \sqrt 2 } \right)}}{3} + \dfrac{{4.\left( {\sqrt 6  - \sqrt 2 } \right)}}{4} + \dfrac{{\sqrt 6  - \sqrt 5 }}{1}\\
 = \left( {\sqrt 5  + \sqrt 2 } \right) + \left( {\sqrt 6  - \sqrt 2 } \right) + \left( {\sqrt 6  - \sqrt 5 } \right)\\
 = \sqrt 5  + \sqrt 2  + \sqrt 6  - \sqrt 2  + \sqrt 6  - \sqrt 5 \\
 = 2\sqrt 6 \\
c,\\
\left( {\dfrac{2}{{\sqrt 5  - \sqrt 3 }} - \dfrac{2}{{\sqrt 5  + \sqrt 3 }}} \right):\dfrac{{2 + \sqrt 3 }}{{\sqrt 3  - 2}}\\
 = \dfrac{{2.\left( {\sqrt 5  + \sqrt 3 } \right) - 2.\left( {\sqrt 5  - \sqrt 3 } \right)}}{{\left( {\sqrt 5  - \sqrt 3 } \right)\left( {\sqrt 5  + \sqrt 3 } \right)}}:\dfrac{{{{\left( {\sqrt 3  + 2} \right)}^2}}}{{\left( {\sqrt 3  - 2} \right)\left( {\sqrt 3  + 2} \right)}}\\
 = \dfrac{{2\sqrt 5  + 2\sqrt 3  - 2\sqrt 5  + 2\sqrt 3 }}{{{{\sqrt 5 }^2} - {{\sqrt 3 }^2}}}:\dfrac{{{{\sqrt 3 }^2} + 2.\sqrt 3 .2 + {2^2}}}{{{{\sqrt 3 }^2} - {2^2}}}\\
 = \dfrac{{4\sqrt 3 }}{{5 - 3}}:\dfrac{{3 + 4\sqrt 3  + 4}}{{3 - 4}}\\
 = \dfrac{{4\sqrt 3 }}{{ - 2}}:\dfrac{{7 + 4\sqrt 3 }}{{ - 1}}\\
 = \left( { - 2\sqrt 3 } \right):\left[ { - \left( {7 + 4\sqrt 3 } \right)} \right]\\
 = \dfrac{{2\sqrt 3 }}{{7 + 4\sqrt 3 }}\\
 = \dfrac{{2\sqrt 3 .\left( {7 - 4\sqrt 3 } \right)}}{{\left( {7 + 4\sqrt 3 } \right)\left( {7 - 4\sqrt 3 } \right)}}\\
 = \dfrac{{14\sqrt 3  - 2.4.{{\sqrt 3 }^2}}}{{{7^2} - {{\left( {4\sqrt 3 } \right)}^2}}}\\
 = \dfrac{{14\sqrt 3  - 2.4.3}}{{49 - 48}}\\
 = \dfrac{{14\sqrt 3  - 24}}{1}\\
 = 14\sqrt 3  - 24
\end{array}\)