tìm số tự nhiên x,y biết 7(x-2021)^2=23-y^2 24/09/2021 Bởi Alice tìm số tự nhiên x,y biết 7(x-2021)^2=23-y^2
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 $\left( {x;y} \right) \in \left\{ {\left( {2022;4} \right);\left( {2020;4} \right)} \right\}$ Bình luận
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 $\left( {x;y} \right) \in \left\{ {\left( {2022;4} \right);\left( {2020;4} \right)} \right\}$