ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 1286

\[\boxed{\text{1286.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[x^{2} - y^{2} = 69\]

\[(x - y)(x + y) = 69,\]

\[\text{\ \ }при\ \text{x\ }и\ y\ \in N:\]

\[\left\{ \begin{matrix} (x - y)(x + y) = 69 \cdot 1 \\ (x - y)(x + y) = 23 \cdot 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} x + y = 69 \\ x - y = 1 \\ \end{matrix} + \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x + y = 1 \\ x - y = 69 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x + y = 23 \\ x - y = 3 \\ \end{matrix} + \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x + y = 3 \\ x - y = 23 \\ \end{matrix} \right.\ \]

\(x + y + x - y = 70\) \(корней\ \)

\(нет\) \(x + y + x - y = 26\)

\(корней\ нет\)

\(2x = 70\)

\(2x = 26\)

\(\left\{ \begin{matrix} x = 35 \\ y = 34 \\ \end{matrix} \right.\ \)

\(\left\{ \begin{matrix} x = 13 \\ y = 10 \\ \end{matrix} \right.\ \)

\(Ответ:(35;34),\ \ (13;10).\)