Решебник по алгебре 9 класс Макарычев Задание 473

\[\boxed{\text{473\ (473).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[первого\ туриста,\]

\[\ а\ \text{y\ }\frac{км}{ч} - второго.\ \]

\[Скорость\ одного\ на\ 1\ \frac{км}{ч}\ \]

\[меньше,\ чем\ скорость\ другого:\]

\[x = y + 1.\]

\[Первый\ двигался\ до\ пункта\]

\[\ \frac{18}{x}\ ч,\ второй - \frac{18}{y}\ ч.\]

\[При\ этом\ первый\ затратил\ на\ \]

\[дорогу\ на\ 54\ мин = \frac{9}{10}\ ч\ \]

\[больше,\ чем\]

\[второй:\]

\[\frac{18}{x} - \frac{18}{y} = \frac{9}{10}.\]

\[Составим\ систему\ уравнений:\]

\[\left\{ \begin{matrix} x = y + 1\ \ \ \ \ \ \ \\ \frac{18}{x} + \frac{9}{10} = \frac{18}{y} \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{2^{\backslash 10y}}{y + 1} + \frac{1^{\backslash y(y + 1)}}{10} = \frac{2^{10(y + 1)}}{y} \\ \end{matrix} \right.\ \Longrightarrow\]

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

\[D = 1 + 4 \cdot 20 = 81\]

\[y_{1,2} = \frac{- 1 \pm 9}{2} = 4;\ - 5.\]

\[Так\ как\ y > 0:\]

\[y = 4 \Longrightarrow x = 5.\]

\[Ответ:4\ \frac{км}{ч}\ и\ 5\ \frac{км}{ч}\text{.\ }\]

\[\boxed{\text{473.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + xy - 2y^{2} - x + y = 0 \\ x^{2} + y^{2} = 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (x + 2y - 1)(x - y) = 0 \\ x² + y² = 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ \left\{ \begin{matrix} x + 2y - 1 = 0 \\ x^{2} + y^{2} = 8\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 1 - 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \\ (1 - 2y)^{2} + y^{2} = 8 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 1 - 2y\ \ \ \ \ \ \ \ \ \ \ \ \\ 5y² - 4y - 7 = 0 \\ \end{matrix} \right.\ \]

\[5y^{2} - 4y - 7 = 0\]

\[D = 4 + 35 = 39\]

\[y_{1,2} = \frac{2 \pm \sqrt{39}}{5};\]

\[\left\{ \begin{matrix} y_{1} = \frac{2 + \sqrt{39}}{5} \\ x_{1} = \frac{1 - 2\sqrt{39}}{5}\ \\ \end{matrix} \right.\ \text{\ \ }или\]

\[\text{\ \ }\left\{ \begin{matrix} y_{2} = \frac{2 - \sqrt{39}}{5} \\ x_{2} = \frac{1 + 2\sqrt{39}}{5}. \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} x - y = 0\ \ \ \\ x^{2} + y^{2} = 8 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y\ \ \ \ \ \ \ \ \ \ \\ x^{2} + x^{2} = 8 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = y \\ x^{2} = 4 \\ \end{matrix} \right.\ \Longrightarrow\]

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

\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} - 6xy + 5y^{2} - x + 5y = 0 \\ x^{2} - 20y^{2} = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (x - y - 1)(x - 5y) = 0 \\ x² - 20y^{2} = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ \left\{ \begin{matrix} x - y - 1 = 0 \\ x^{2} - {20y}^{2} = 5 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + 2y + 1 - 20y^{2} = 5 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 19y² - 2y + 4 = 0 \\ \end{matrix} \right.\ \]

\[19y^{2} - 2y + 4 = 0\]

\[D = 1 - 19 \cdot 4 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет;\]

\[2)\ \left\{ \begin{matrix} x - 5y = 0\ \ \ \ \ \\ x^{2} - 20y^{2} = 5 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 5y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 25y^{2} - 20y^{2} = 5 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 5y \\ y^{2} = 1\ \ \\ \end{matrix} \right.\ \Longrightarrow\]

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

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

\[Ответ:а)\ ( - 2;\ - 2);(2;2);\]

\[\left( \frac{1 + 2\sqrt{39}}{5};\ \frac{2 - \sqrt{39}}{5} \right);\]

\[\left( \frac{1 - 2\sqrt{39}}{5};\ \frac{2 + \sqrt{39}}{5} \right);\]

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