Đáp án:
$\begin{array}{l}
1)\sqrt {15} + \sqrt {17} - 8\\
= \sqrt {17} - 4 - \left( {4 - \sqrt {15} } \right)\\
= \dfrac{{17 - 16}}{{\sqrt {17} + 4}} - \dfrac{{16 - 15}}{{4 + \sqrt {15} }}\\
= \dfrac{1}{{\sqrt {17} + 4}} - \dfrac{1}{{\sqrt {15} + 4}}\\
Do:\sqrt {17} + 4 > \sqrt {15} + 4\\
\Leftrightarrow \dfrac{1}{{\sqrt {17} + 4}} < \dfrac{1}{{\sqrt {15} + 4}}\\
\Leftrightarrow \dfrac{1}{{\sqrt {17} + 4}} - \dfrac{1}{{\sqrt {15} + 4}} < 0\\
\Leftrightarrow \sqrt {15} + \sqrt {17} - 8 < 0\\
\Leftrightarrow \sqrt {15} + \sqrt {17} < 8\\
2)\sqrt {2020} + \sqrt {2022} - 2\sqrt {2021} \\
= \sqrt {2022} - \sqrt {2021} - \left( {\sqrt {2021} - \sqrt {2020} } \right)\\
= \dfrac{1}{{\sqrt {2020} + \sqrt {2021} }} - \dfrac{1}{{\sqrt {2021} + \sqrt {2020} }} < 0\\
\Leftrightarrow \sqrt {2020} + \sqrt {2022} - 2\sqrt {2021} < 0\\
\Leftrightarrow \sqrt {2020} + \sqrt {2022} < 2\sqrt {2021} \\
3)\sqrt {2022} - \sqrt {2021} = \dfrac{1}{{\sqrt {2022} + \sqrt {2021} }}\\
\sqrt {2021} - \sqrt {2019} = \dfrac{2}{{\sqrt {2021} + \sqrt {2019} }} = \dfrac{1}{{\dfrac{{\sqrt {2021} + \sqrt {2019} }}{2}}}\\
Do:\dfrac{1}{{\sqrt {2022} + \sqrt {2021} }} < \dfrac{1}{{\dfrac{{\sqrt {2021} + \sqrt {2019} }}{2}}}\\
\Leftrightarrow \sqrt {2022} - \sqrt {2021} < \sqrt {2021} - \sqrt {2019}
\end{array}$
