Đáp án:
$\sqrt {2021} - \sqrt {2020} > \sqrt {2022} - \sqrt {2021} $
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\sqrt {2021} - \sqrt {2020} = \dfrac{{{{\left( {\sqrt {2021} } \right)}^2} - {{\left( {\sqrt {2020} } \right)}^2}}}{{\sqrt {2021} + \sqrt {2020} }} = \dfrac{1}{{\sqrt {2021} + \sqrt {2020} }}\\
\sqrt {2022} - \sqrt {2021} = \dfrac{{{{\left( {\sqrt {2022} } \right)}^2} - {{\left( {\sqrt {2021} } \right)}^2}}}{{\sqrt {2022} + \sqrt {2021} }} = \dfrac{1}{{\sqrt {2022} + \sqrt {2021} }}
\end{array}$
Mà $0 < \sqrt {2021} + \sqrt {2020} < \sqrt {2022} + \sqrt {2021} $
$\begin{array}{l}
\Rightarrow \dfrac{1}{{\sqrt {2021} + \sqrt {2020} }} > \dfrac{1}{{\sqrt {2022} + \sqrt {2021} }}\\
\Rightarrow \sqrt {2021} - \sqrt {2020} > \sqrt {2022} - \sqrt {2021}
\end{array}$
Vậy$\sqrt {2021} - \sqrt {2020} > \sqrt {2022} - \sqrt {2021} $
