Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 360

\[\boxed{\mathbf{360}.}\]

\[1)\ \left\{ \begin{matrix} xy - x + y = 7\ \ \ \\ xy + x - y = 13 \\ \end{matrix} \right.\ ( + )\backslash\text{(} - )\]

\[\left\{ \begin{matrix} 2xy = 20\ \ \ \ \ \ \ \ \ \ \ \ \\ - 2x + 2y = - 6 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} xy = 10\ \ \ \\ x - y = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} x = 3 + y\ \ \ \ \ \ \ \ \ \\ (3 + y)y = 10 \\ \end{matrix} \right.\ \]

\[y^{2} + 3y - 10 = 0\]

\[y_{1} + y_{2} = - 3;\ \ y_{1} \cdot y_{2} = - 10\]

\[y_{1} = - 5;\ \ \ \ \ \ \ y_{2} = 2.\]

\[x_{1} = - 2;\ \ \ \ \ \ \ x_{2} = 5.\]

\[Ответ:( - 2;\ - 5);(5;2).\]

\[2)\ \left\{ \begin{matrix} xy - 2 \cdot (x + y) = 2 \\ xy + x + y = 29\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x + y = 29 - xy\ \ \ \ \ \ \ \ \ \ \ \\ xy - 2 \cdot (29 - xy) = 2 \\ \end{matrix} \right.\ \]

\[xy - 58 + 2xy = 2\]

\[3xy = 60\]

\[xy = 20.\]

\[x + y = 29 - xy\]

\[x + y = 29 - 20\]

\[x + y = 9.\]

\[\left\{ \begin{matrix} x + y = 9 \\ xy = 20\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x = 9 - y\ \ \ \ \ \ \ \ \\ (9 - y)y = 20 \\ \end{matrix} \right.\ \]

\[y^{2} - 9y + 20 = 0\]

\[y_{1} + y_{2} = 9;\ \ \ y_{1} \cdot y_{2} = 20\]

\[y_{1} = 5;\ \ \ y_{2} = 4\]

\[x_{1} = 4;\ \ \ x_{2} = 5.\]

\[Ответ:(4;5);(5;4).\]