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МатематикаРешебник по алгебре 9 класс Макарычев Задание 472
Задание 472
\[\boxed{\text{472\ (472).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[Пусть\ \text{x\ }\frac{км}{ч} - скорость\]
\[\ первого\ пешехода;\ \]
\[\text{y\ }\frac{км}{ч} - скорость\ второго.\]
\[Спустя\ 4\ часа\ они\ пройдут:\]
\[4x + 4y + 4 = 40\]
\[4 \cdot (x + y) = 36\]
\[x + y = 9.\]
\[До\ середины\ пути\ второй\ \]
\[пешеход\ мог\ дойти\ за\ \frac{20}{y}\ ч,\ \]
\[первый - за\ \frac{20}{x}\ ч.\ Первый\ бы\]
\[\ затратил\ на\ дорогу\ \]
\[на\ 1\ ч\ больше:\]
\[\frac{20}{x} - \frac{20}{y} = 1\]
\[Составим\ систему\ уравнений6\ \]
\[\left\{ \begin{matrix} x + y = 9\ \ \ \ \ \\ \frac{20}{x} - \frac{20}{y} = 1 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 9 - x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{20^{\backslash 9 - x}}{x} - \frac{20^{\backslash x}}{9 - x} = 1^{\backslash x(9 - x)} \\ \end{matrix} \Longrightarrow \right.\ \]
\[\Longrightarrow \left\{ \begin{matrix} y = 9 - x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 180 - 20x - 20x - 9x + x^{2} = 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 49x + 180 = 0\]
\[x_{1} + x_{2} = 49;\ \ \ x_{1} \cdot x_{2} = 180\]
\[x_{1} = 4;\ \ \ \ \ \ x_{2} = 45;\]
\[\left\{ \begin{matrix} x_{1} = 4 \\ y_{1} = 5 \\ \end{matrix} \right.\ \ \ или\ \ \left\{ \begin{matrix} x_{2} = 45\ \ \ \ \\ y_{2} = - 36 \\ \end{matrix} \right.\ ;\ \ \]
\[но\ y > 0 \Longrightarrow не\ подходит.\]
\[Ответ:\ 4\ \ \frac{км}{ч}\ \ и\ \ 5\ \ \frac{км}{ч}.\]
\[\boxed{\text{472.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} (x - 2y)(x + 3y) = 0 \\ x² - y^{2} = 12\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[1)\ \left\{ \begin{matrix} x - 2y = 0\ \ \ \\ x^{2} - y^{2} = 12 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4y^{2} - y^{2} = 12 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 2y \\ y^{2} = 4 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = \pm 2 \\ x = \pm 4 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 2 \\ x_{1} = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ }или\ \ \ \left\{ \begin{matrix} y_{2} = - 2 \\ x_{2} = - 4 \\ \end{matrix} \right.\ ;\]
\[2)\ \left\{ \begin{matrix} x + 3y = 0\ \ \\ x^{2} - y^{2} = 12 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - 3y\ \ \ \ \ \ \ \ \ \\ 9y^{2} - y^{2} = 12 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - 3y \\ y^{2} = \frac{3}{2}\text{\ \ \ } \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = \pm \frac{\sqrt{6}}{2} \\ x = - 3y \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = \frac{\sqrt{6}}{2}\text{\ \ \ \ \ \ \ } \\ x_{1} = - \frac{3\sqrt{6}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \]
\[\left\{ \begin{matrix} y_{2} = - \frac{\sqrt{6}}{2} \\ x_{2} = \frac{3\sqrt{6}}{2}. \\ \end{matrix} \right.\ \]
\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} - 4xy + 3y^{2} + 2x - 6y = 0 \\ x^{2} - xy + y^{2} = 7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} (x - y + 2)(x - 3y) = 0 \\ x² - xy + y^{2} = 7\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[1)\ \left\{ \begin{matrix} x - y + 2 = 0\ \ \ \ \ \\ x^{2} - xy + y^{2} = 7 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = x + 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - x(x + 2) + (x + 2)^{2} = 7 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = x + 2\ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2x - 3 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x_{1} = 1 \\ y_{1} = 3 \\ \end{matrix} \right.\ \text{\ \ }или\ \ \left\{ \begin{matrix} x_{2} = - 3 \\ y_{2} = - 1. \\ \end{matrix} \right.\ \]
\[2)\ \left\{ \begin{matrix} x - 3y = 0\ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - xy + y^{2} = 7 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 3y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 9y^{2} - 3y^{2} + y^{2} = 7 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 3y \\ y^{2} = 1\ \\ \end{matrix} \Longrightarrow \right.\ \]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 1 \\ x_{1} = 3 \\ \end{matrix} \right.\ \ \ или\ \left\{ \begin{matrix} y_{2} = - 1 \\ x_{2} = - 3. \\ \end{matrix} \right.\ \]
\[Ответ:а)\ (4;2);( - 4;\ - 2);\]
\[\left( - \frac{3\sqrt{6}}{2};\frac{\sqrt{6}}{2} \right);\left( \frac{3\sqrt{6}}{2};\ - \frac{\sqrt{6}}{2} \right);\ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ б)\ (1;3);( - 3;\ - 1);(3;1).\]
