Đáp án:
$\begin{array}{l}
32)\sqrt 2 - \sqrt 6 \\
= \sqrt 2 .\left( {1 - \sqrt 3 } \right)\\
Do:1 - \sqrt 3 < 0\\
\Leftrightarrow \sqrt 2 \left( {1 - \sqrt 3 } \right) < 1 - \sqrt 3 \\
\Leftrightarrow 1 - \sqrt 3 > \sqrt 2 - \sqrt 6 \\
33)2 + \sqrt 2 - \left( {5 - \sqrt 3 } \right)\\
= \sqrt 3 - \sqrt 2 - 3 < 0\\
\Leftrightarrow 2 + \sqrt 2 < 5 - \sqrt 3 \\
34)\\
\sqrt {4\sqrt 5 } = \sqrt {\sqrt {16.5} } = \sqrt {\sqrt {80} } \\
\sqrt {5\sqrt 3 } = \sqrt {\sqrt {25.3} } = \sqrt {\sqrt {75} } < \sqrt {\sqrt {80} } \\
\Leftrightarrow \sqrt {4\sqrt 5 } > \sqrt {5\sqrt 3 } \\
35)\sqrt {\sqrt 6 - \sqrt 5 } - \sqrt {\sqrt 3 - \sqrt 2 } \\
= \sqrt {\dfrac{1}{{\sqrt 6 + \sqrt 5 }}} - \sqrt {\dfrac{1}{{\sqrt 3 + \sqrt 2 }}} < 0\\
36) - 2\sqrt {\dfrac{1}{2}\sqrt 5 } = - \sqrt {4.\dfrac{1}{2}\sqrt 5 } = - \sqrt {\sqrt {20} } \\
- 3\sqrt {\dfrac{1}{3}\sqrt 2 } = - \sqrt {9.\dfrac{1}{3}\sqrt 2 } = - \sqrt {\sqrt {18} } > - \sqrt {\sqrt {20} } \\
\Leftrightarrow - 2\sqrt {\dfrac{1}{2}\sqrt 5 } < - 3\sqrt {\dfrac{1}{3}\sqrt 2 } \\
37)\\
\sqrt {2003} + \sqrt {2005} - 2\sqrt {2004} \\
= \sqrt {2005} - \sqrt {2004} - \left( {\sqrt {2004} - \sqrt {2003} } \right)\\
= \dfrac{{2005 - 2004}}{{\sqrt {2005} + \sqrt {2004} }} - \dfrac{{2004 - 2003}}{{\sqrt {2004} + \sqrt {2003} }}\\
= \dfrac{1}{{\sqrt {2005} + \sqrt {2004} }} - \dfrac{1}{{\sqrt {2004} + \sqrt {2003} }} < 0\\
\Leftrightarrow \sqrt {2003} + \sqrt {2005} < 2\sqrt {2004}
\end{array}$
