Đáp án:
$\begin{array}{l}
+ \sqrt {49} = 7 = 4 + 3 = \sqrt {16} + \sqrt 9 > \sqrt 7 + \sqrt 5 \\
\Leftrightarrow \sqrt 7 + \sqrt 5 < \sqrt {49} \\
+ \sqrt 2 + \sqrt {11} < \sqrt 3 + \sqrt {25} \\
\Leftrightarrow \sqrt 2 + \sqrt {11} < \sqrt 3 + 5\\
+ \dfrac{1}{2}\sqrt {\dfrac{{17}}{2}} = \dfrac{{\sqrt {34} }}{4} = \dfrac{{\sqrt {306} }}{{12}}\\
\dfrac{1}{3}\sqrt {19} = \dfrac{{\sqrt {19.16} }}{{12}} = \dfrac{{\sqrt {304} }}{{12}} < \dfrac{{\sqrt {306} }}{{12}}\\
\Leftrightarrow \dfrac{1}{2}\sqrt {\dfrac{{17}}{2}} > \dfrac{1}{3}\sqrt {19} \\
+ )\sqrt {21} - \sqrt 5 - \left( {\sqrt {20} - \sqrt 6 } \right)\\
= \sqrt {21} - \sqrt {20} + \sqrt 6 - \sqrt 5 > 0\\
\Leftrightarrow \sqrt {21} - \sqrt 5 > \sqrt {20} - \sqrt 6 \\
+ )\dfrac{1}{4}\sqrt {82} = \dfrac{{\sqrt {82.7} }}{{4\sqrt 7 }} = \dfrac{{\sqrt {574} }}{{4\sqrt 7 }}\\
6\sqrt {\dfrac{1}{7}} = \dfrac{{6.4}}{{4\sqrt 7 }} = \dfrac{{\sqrt {{{24}^2}} }}{{4\sqrt 7 }} = \dfrac{{\sqrt {576} }}{{4\sqrt 7 }}\\
\Leftrightarrow \dfrac{1}{4}\sqrt {82} < 6\sqrt {\dfrac{1}{7}} \\
+ )Do:\sqrt 6 + \sqrt {20} > 1 + \sqrt 5 \\
\Leftrightarrow \sqrt {\sqrt 6 + \sqrt {20} } > \sqrt {1 + \sqrt 5 }
\end{array}$
