Đáp án:
1) \(\sqrt {2019} + \sqrt {2021} < 2\sqrt {2020} \)
2)
Giải thích các bước giải:
\(\begin{array}{l}
1.Có:\\
{\left( {\sqrt {2019} + \sqrt {2021} } \right)^2}\\
= 2019 + 2021 + 2\sqrt {2019.2021} \\
= 2020 + 2020 + 2\sqrt {\left( {2020 - 1} \right)\left( {2020 + 1} \right)} \\
= 2.2020 + 2\sqrt {{{2020}^2} - 1} < 2.2020 + 2.2020\\
\to 2.2020 + 2\sqrt {{{2020}^2} - 1} < 4.2020\\
\to 2.2020 + 2\sqrt {{{2020}^2} - 1} < {\left( {2\sqrt {2020} } \right)^2}\\
\to \sqrt {2019} + \sqrt {2021} < 2\sqrt {2020}
\end{array}\)
\(\begin{array}{l}
2)\sqrt {2 + 2\sqrt 2 + 1} + \sqrt {{{\left( {3 - \sqrt 2 } \right)}^2}} = 4\\
\Leftrightarrow \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} + \sqrt {{{\left( {3 - \sqrt 2 } \right)}^2}} = 4\\
\Leftrightarrow \sqrt 2 + 1 + 3 - \sqrt 2 = 4\\
\Leftrightarrow 4 = 4\left( {ld} \right)\\
\to dpcm
\end{array}\)
