Đáp án:
$\begin{array}{l}
4)\sqrt {13} + \sqrt {11} - 4\sqrt 3 \\
= \sqrt {13} - 2\sqrt 3 - \left( {2\sqrt 3 - \sqrt {11} } \right)\\
= \sqrt {13} - \sqrt {12} - \left( {\sqrt {12} - \sqrt {11} } \right)\\
= \dfrac{{13 - 12}}{{\sqrt {13} + \sqrt {12} }} - \dfrac{{12 - 11}}{{\sqrt {12} + \sqrt {11} }}\\
= \dfrac{1}{{\sqrt {13} + \sqrt {12} }} - \dfrac{1}{{\sqrt {12} + \sqrt {11} }}\\
Do:\sqrt {13} + \sqrt {12} > \sqrt {12} + \sqrt {11} \\
\Leftrightarrow \dfrac{1}{{\sqrt {13} + \sqrt {12} }} < \dfrac{1}{{\sqrt {12} + \sqrt {11} }}\\
\Leftrightarrow \dfrac{1}{{\sqrt {13} + \sqrt {12} }} - \dfrac{1}{{\sqrt {12} + \sqrt {11} }} < 0\\
\Leftrightarrow \sqrt {13} + \sqrt {11} - 4\sqrt 3 < 0\\
\Leftrightarrow \sqrt {13} + \sqrt {11} < 4\sqrt 3 \\
5)\dfrac{{15}}{{\sqrt {14} }} + \dfrac{{14}}{{\sqrt {15} }} - \left( {\sqrt {14} + \sqrt {15} } \right)\\
= \dfrac{{15}}{{\sqrt {14} }} - \sqrt {14} + \dfrac{{14}}{{\sqrt {15} }} - \sqrt {15} \\
= \dfrac{{15 - 14}}{{\sqrt {14} }} + \dfrac{{14 - 15}}{{\sqrt {15} }}\\
= \dfrac{1}{{\sqrt {14} }} - \dfrac{1}{{\sqrt {15} }} > 0\\
\Leftrightarrow \dfrac{{15}}{{\sqrt {14} }} + \dfrac{{14}}{{\sqrt {15} }} - \left( {\sqrt {14} + \sqrt {15} } \right) > 0\\
\Leftrightarrow \dfrac{{15}}{{\sqrt {14} }} + \dfrac{{14}}{{\sqrt {15} }} > \sqrt {14} + \sqrt {15}
\end{array}$
