ГДЗ по алгебре 9 класс Макарычев Задание 513

\[\boxed{\text{513\ (513).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + y^{2} = 25 \\ xy = 12\ \ \ \ \ \ \ \ \\ \end{matrix} \Longrightarrow \right.\ \]

\[\Longrightarrow \left\{ \begin{matrix} x^{2} + y^{2} = 25 \\ 2xy = 24\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x^{2} + y^{2} - 2xy = 25 - 24 \\ xy = 12\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x^{2} - 2xy + y^{2} = 1 \\ xy = 12\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (x - y)^{2} = 1 \\ xy = 12\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x - y = \pm 1 \\ xy = 12\ \ \ \ \ \ \ \\ \end{matrix} \right.\ ;\]

\[1)\ \left\{ \begin{matrix} x - y = 1 \\ xy = 12\ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 1\ \ \ \ \ \ \ \\ y(y + 1) = 12 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 1\ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + y - 12 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y_{1} = - 4 \\ x_{1} = - 3 \\ \end{matrix} \right.\ \ \ \ \ или\ \left\{ \begin{matrix} y_{2} = 3 \\ x_{2} = 4. \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} x - y = - 1 \\ xy = 12\ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y - 1\ \ \ \ \ \ \\ y(y - 1) = 12 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y - 1\ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - y - 12 = 0 \\ \end{matrix} \right.\ \]

\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 4 \\ x_{1} = 3 \\ \end{matrix} \right.\ \ \ \ или\ \ \ \left\{ \begin{matrix} y_{2} = - 3 \\ x_{2} = - 4. \\ \end{matrix} \right.\ \]

\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} + y^{2} = 26 \\ x + y = 6\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 6 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 36 - 12y + y^{2} + y^{2} = 26 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 6 - y\ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 6y + 5 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 1 \\ x_{1} = 5 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }или\ \left\{ \begin{matrix} y_{2} = 5 \\ x_{2} = 1. \\ \end{matrix} \right.\ \]

\[Ответ:а)\ ( - 3;\ - 4);(4;3);(3;4);\]

\[( - 4;\ - 3);\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ (5;1);(1;5).\]