Đáp án:
$\begin{array}{l}
a)\dfrac{{1 + \sqrt 2 }}{{1 - \sqrt 2 }} = \dfrac{{{{\left( {1 + \sqrt 2 } \right)}^2}}}{{\left( {1 - \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)}} = \dfrac{{3 + 2\sqrt 2 }}{{1 - 2}} = - 3 - 2\sqrt 2 \\
b)\dfrac{{\sqrt {2 + \sqrt 3 } }}{{\sqrt {2 - \sqrt 3 } }} = \dfrac{{\sqrt {4 + 2\sqrt 3 } }}{{\sqrt {4 - 2\sqrt 3 } }}\\
= \dfrac{{\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }}{{\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} }} = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}\\
= \dfrac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{{3 - 1}} = \dfrac{{4 + 2\sqrt 3 }}{2} = 2 + \sqrt 3 \\
c)\dfrac{{1 - {a^2}}}{{1 - \sqrt a }} = \dfrac{{\left( {1 - a} \right)\left( {1 + a} \right)}}{{1 - \sqrt a }}\\
= \dfrac{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a } \right)\left( {1 + a} \right)}}{{1 - a}}\\
= \left( {1 + \sqrt a } \right)\left( {1 + a} \right)\\
= 1 + a + \sqrt a + a\sqrt a \\
d)\sqrt {\dfrac{2}{{3 - \sqrt 5 }}} = \sqrt {\dfrac{4}{{6 - 2\sqrt 5 }}} = \dfrac{2}{{\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} }}\\
= \dfrac{2}{{\sqrt 5 - 1}}\\
= \dfrac{{2\left( {\sqrt 5 + 1} \right)}}{{5 - 1}} = \dfrac{{\sqrt 5 + 1}}{2}\\
e)\sqrt {\dfrac{{a - 4}}{{2\left( {\sqrt a - 2} \right)}}} = \sqrt {\dfrac{{\left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}{{2\left( {\sqrt a - 2} \right)}}} \\
= \dfrac{{\sqrt {\sqrt a + 2} }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {2\sqrt a + 4} }}{2}\\
f)\dfrac{{1 + \sqrt 3 }}{{\sqrt 3 - 1}} = \dfrac{{{{\left( {1 + \sqrt 3 } \right)}^2}}}{{3 - 1}} = 2 + \sqrt 3
\end{array}$
