Đáp án:
$\left( {x;y} \right) \in \left\{ {\left( {2022;4} \right);\left( {2020;4} \right)} \right\}$
Giải thích các bước giải:
Ta có:
$7{\left( {x - 2021} \right)^2} = 23 - {y^2}$
Nhận xét:
$\begin{array}{l}
{y^2} \ge 0,\forall y\\
\Rightarrow 23 - {y^2} \le 23\\
\Rightarrow 7{\left( {x - 2021} \right)^2} \le 23\\
\Rightarrow \left[ \begin{array}{l}
{\left( {x - 2021} \right)^2} = 0\\
{\left( {x - 2021} \right)^2} = 1
\end{array} \right.\left( {Do:x \in N} \right)\\
+ )TH1:{\left( {x - 2021} \right)^2} = 0 \Rightarrow x = 2021\\
\Rightarrow 23 - {y^2} = 0\\
\Rightarrow y = \pm \sqrt {23} \left( l \right)\\
+ )TH2:{\left( {x - 2021} \right)^2} = 1\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2021 = 1\\
x - 2021 = - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2022\\
x = 2020
\end{array} \right.
\end{array}$
Mặt khác:
$\begin{array}{l}
{\left( {x - 2021} \right)^2} = 1 \Rightarrow {y^2} = 23 - 7{\left( {x - 2021} \right)^2} = 16\\
\Rightarrow \left[ \begin{array}{l}
y = - 4\left( l \right)\\
y = 4\left( c \right)
\end{array} \right.\\
\Rightarrow y = 4
\end{array}$
Vậy $\left( {x;y} \right) \in \left\{ {\left( {2022;4} \right);\left( {2020;4} \right)} \right\}$
