Đáp án:
\[\left( {x;y} \right) = \left\{ {\left( {53;52} \right);\left( {19;16} \right);\left( {13;8} \right);\left( {11;4} \right)} \right\}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{x^2} - {y^2} = 105\\
\Leftrightarrow \left( {x - y} \right)\left( {x + y} \right) = 105\\
x,y \in {Z^ + } \Rightarrow \left\{ \begin{array}{l}
\left( {x + y} \right) \in {Z^ + }\\
x + y \ge 2\\
x + y > x - y
\end{array} \right.\\
105 = 1.105 = 3.35 = 5.21 = 7.15\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + y = 105\\
x - y = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x + y = 35\\
x - y = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x + y = 21\\
x - y = 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x + y = 15\\
x - y = 7
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 53\\
y = 52
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 19\\
y = 16
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 13\\
y = 8
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 11\\
y = 4
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left( {x;y} \right) = \left\{ {\left( {53;52} \right);\left( {19;16} \right);\left( {13;8} \right);\left( {11;4} \right)} \right\}
\end{array}\)
