Решебник по алгебре 9 класс Макарычев Задание 513

 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[\boxed{\text{513.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[Пусть\ \text{x\ }\frac{км}{ч}\ \ и\ y\frac{\ км}{ч} -\]

\[скорости\ автомобилей.\]

\[1\ ч\ 15\ мин = \frac{75}{60}\ \ ч = \frac{5}{4}\ ч;\ \ \]

\[\ 48\ мин = \frac{48}{60}\ ч = \frac{4}{5}\ ч.\]

\[Автомобили\ встретятся\ через\ \]

\[\text{\ \ }\frac{90}{x + y}\ ч.\ \]

\[Составим\ систему\ уравнений:\]

\[\left\{ \begin{matrix} 90 - \frac{90}{x + y} \cdot x = \frac{5}{4}x \\ 90 - \frac{90}{x + y} \cdot y = \frac{4}{5}y \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} \frac{90y}{x(x + y)} = \frac{5}{4} \\ \frac{90x}{y(x + y)} = \frac{4}{5} \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \frac{90y}{x(x + y)} = \frac{y(x + y)}{90x} \Longrightarrow\]

\[\Longrightarrow \frac{90}{x + y} = \frac{x + y}{90} \Longrightarrow\]

\[\Longrightarrow x + y = 90 \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 90 - x = \frac{5}{4}x \\ 90 - y = \frac{4}{5}y \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} \frac{9}{4}x = 90 \\ \frac{9}{5}y = 90 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = 40 \\ y = 50 \\ \end{matrix} \right.\ .\]

\[Ответ:40\ \frac{км}{ч}\ \ и\ 50\ \frac{км}{ч}.\]