Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sqrt 3 - \sqrt 2 = \dfrac{{\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}{{\sqrt 3 + \sqrt 2 }} = \dfrac{{3 - 2}}{{\sqrt 3 + \sqrt 2 }} = \dfrac{1}{{\sqrt 3 + \sqrt 2 }}\\
\sqrt 4 - \sqrt 3 = \dfrac{{\left( {\sqrt 4 - \sqrt 3 } \right)\left( {\sqrt 4 + \sqrt 3 } \right)}}{{\sqrt 4 + \sqrt 3 }} = \dfrac{{4 - 3}}{{\sqrt 4 + \sqrt 3 }} = \dfrac{1}{{\sqrt 4 + \sqrt 3 }}\\
\sqrt 2 < \sqrt 4 \Rightarrow \sqrt 3 + \sqrt 2 < \sqrt 4 + \sqrt 3 \\
\Rightarrow \dfrac{1}{{\sqrt 3 + \sqrt 2 }} > \dfrac{1}{{\sqrt 4 + \sqrt 3 }}\\
\Leftrightarrow \sqrt 3 - \sqrt 2 > \sqrt 4 - \sqrt 3 \\
b,\\
\sqrt {2021} - \sqrt {2020} = \dfrac{{\left( {\sqrt {2021} - \sqrt {2020} } \right)\left( {\sqrt {2021} + \sqrt {2020} } \right)}}{{\sqrt {2021} + \sqrt {2020} }} = \dfrac{{2021 - 2020}}{{\sqrt {2021} + \sqrt {2020} }} = \dfrac{1}{{\sqrt {2021} + \sqrt {2020} }}\\
\sqrt {2022} - \sqrt {2021} = \dfrac{{\left( {\sqrt {2022} - \sqrt {2021} } \right)\left( {\sqrt {2022} + \sqrt {2021} } \right)}}{{\sqrt {2022} + \sqrt {2021} }} = \dfrac{{2022 - 2021}}{{\sqrt {2022} + \sqrt {2021} }} = \dfrac{1}{{\sqrt {2022} + \sqrt {2021} }}\\
\sqrt {2020} < \sqrt {2022} \Rightarrow \sqrt {2021} + \sqrt {2020} < \sqrt {2021} + \sqrt {2022} \\
\Rightarrow \dfrac{1}{{\sqrt {2021} + \sqrt {2020} }} > \dfrac{1}{{\sqrt {2021} + \sqrt {2022} }}\\
\Leftrightarrow \sqrt {2021} - \sqrt {2020} > \sqrt {2022} - \sqrt {2021}
\end{array}\)
