So sánh căn 2022- căn 2021 và căn 2021- căn 2022

so sánh căn 2021- căn 2020 ѵà căn 2022- căn 2021

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} $

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} $