Đáp án:
$\begin{array}{l}
2)a)\left( {\dfrac{1}{{3 - \sqrt 5 }} - \dfrac{1}{{3 + \sqrt 5 }}} \right):\dfrac{{5 - \sqrt 5 }}{{\sqrt 5 - 1}}\\
= \dfrac{{3 + \sqrt 5 - \left( {3 - \sqrt 5 } \right)}}{{9 - 5}}.\dfrac{{\sqrt 5 - 1}}{{\sqrt 5 \left( {\sqrt 5 - 1} \right)}}\\
= \dfrac{{2\sqrt 5 }}{4}.\dfrac{1}{{\sqrt 5 }}\\
= \dfrac{1}{2}\\
b)\left( {\dfrac{{3 + 2\sqrt 3 }}{{\sqrt 3 }} + \dfrac{{2 + \sqrt 2 }}{{\sqrt 2 + 1}}} \right) - \dfrac{1}{{\sqrt 3 + \sqrt 2 }}\\
= \left( {\sqrt 3 + 2 + \sqrt 2 } \right) - \dfrac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}}\\
= 2 + \sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 \\
= 2 + 2\sqrt 2 \\
c)\dfrac{{\sqrt {3 - \sqrt 5 } \left( {3 + \sqrt 5 } \right)}}{{\sqrt {10} + \sqrt 2 }}\\
= \dfrac{{\sqrt {\left( {3 - \sqrt 5 } \right)\left( {3 + \sqrt 5 } \right)} .\sqrt {3 + \sqrt 5 } }}{{\sqrt 2 \left( {\sqrt 5 + 1} \right)}}\\
= \dfrac{{\sqrt {9 - 5} .\sqrt {3 + \sqrt 5 } }}{{\sqrt 2 .\left( {\sqrt 5 + 1} \right)}}\\
= \dfrac{{\sqrt 2 .\sqrt 2 .\sqrt {3 + \sqrt 5 } }}{{\sqrt 2 \left( {\sqrt 5 + 1} \right)}}\\
= \dfrac{{\sqrt {6 + 2\sqrt 5 } }}{{\sqrt 5 + 1}}\\
= \dfrac{{\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} }}{{\sqrt 5 + 1}}\\
= \dfrac{{\sqrt 5 + 1}}{{\sqrt 5 + 1}}\\
= 1\\
d)\dfrac{1}{{\sqrt 5 + \sqrt 2 }} + \dfrac{1}{{\sqrt 5 - \sqrt 2 }}\\
= \dfrac{{\sqrt 5 - \sqrt 2 + \sqrt 5 + \sqrt 2 }}{{5 - 2}}\\
= \dfrac{{2\sqrt 5 }}{3}\\
B4)a)\sqrt {10} - \sqrt 7 = \dfrac{{10 - 7}}{{\sqrt {10} + \sqrt 7 }} = \dfrac{3}{{\sqrt {10} + \sqrt 7 }}\\
\sqrt {13} - \sqrt {10} = \dfrac{{13 - 10}}{{\sqrt {13} + \sqrt {10} }} = \dfrac{3}{{\sqrt {13} + \sqrt {10} }}\\
Do:\sqrt {10} + \sqrt 7 < \sqrt {13} + \sqrt {10} \\
\Leftrightarrow \dfrac{3}{{\sqrt {10} + \sqrt 7 }} > \dfrac{3}{{\sqrt {13} + \sqrt {10} }}\\
\Leftrightarrow \sqrt {10} - \sqrt 7 > \sqrt {13} - \sqrt {10} \\
b)\sqrt {2013} - \sqrt {2012} = \dfrac{{2013 - 2012}}{{\sqrt {2013} + \sqrt {2012} }} = \dfrac{1}{{\sqrt {2013} + \sqrt {2012} }}\\
\sqrt {2014} - \sqrt {2013} = \dfrac{1}{{\sqrt {2014} + \sqrt {2013} }}\\
Do:\sqrt {2013} + \sqrt {2012} < \sqrt {2014} + \sqrt {2013} \\
\Leftrightarrow \dfrac{1}{{\sqrt {2013} + \sqrt {2012} }} > \dfrac{1}{{\sqrt {2014} + \sqrt {2013} }}\\
\Leftrightarrow \sqrt {2013} - \sqrt {2012} > \sqrt {2014} - \sqrt {2013}
\end{array}$
