Решебник по алгебре 9 класс Макарычев Задание 436

 

\[\boxed{\text{436\ (436).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \left\{ \begin{matrix} 6 \cdot (x - y) - 50 = y \\ y - xy = 24\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 6y - 6x - 50 - y = 0 \\ y - xy = 24\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 5y - 6x - 50 = 0 \\ y - xy = 24\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 5y = 6x + 50 \\ y - xy = 24\ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 1,2x + 10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 1,2x + 10 - 1,2x^{2} - 10x = 24 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 1,2x + 10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 1,2x^{2} + 8,8x + 14 = 0 \\ \end{matrix} \right.\ \]

\[1,2x^{2} + 8,8x + 14 = 0\ \ \ \ | \cdot 5\]

\[6x^{2} + 44x + 70 = 0\ \ \ \ \ \ \ \ \ |\ :2\]

\[3x^{2} + 22x + 35 = 0\]

\[D_{1} = 11^{2} - 3 \cdot 35 =\]

\[= 121 - 105 = 16\]

\[x_{1,2} = \frac{- 11 \pm 4}{3} = - 5;\ - 2\frac{1}{3}.\]

\[1)\ \left\{ \begin{matrix} x_{1} = - 5 \\ y_{1} = 4\ \ \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} x_{2} = 7\frac{1}{5}\text{\ \ \ \ } \\ y_{2} = - 2\frac{1}{3} \\ \end{matrix} \right.\ .\]

\[\textbf{б)}\ \left\{ \begin{matrix} p + 5t = 2 \cdot (p + t) \\ pt - t = 10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} p + 5t = 2p + 2t \\ pt - t = 10\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} p = 3t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3t^{2} - t - 10 = 0 \\ \end{matrix} \right.\ \]

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

\[D = 1 + 4 \cdot 3 \cdot 10 = 121\]

\[t_{1,2} = \frac{1 \pm 11}{6} = 2;\ - 1\frac{2}{3}.\]

\[1)\ \left\{ \begin{matrix} t_{1} = 2 \\ p_{1} = 6 \\ \end{matrix} \right.\ \text{\ \ }\]

\[2)\ \left\{ \begin{matrix} t_{2} = - 1\frac{2}{3} \\ p_{2} = - 5\ \ \\ \end{matrix} \right.\ .\]

\[Ответ:а)\ ( - 5;4);\left( - 2\frac{1}{3};7\frac{1}{5} \right);\ \ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ (6;2);\ \ \left( - 5;\ - 1\frac{2}{3} \right).\]

\(\boxed{\text{436.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\)

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

\[\ первого\ пешехода;\ \]

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

\[Спустя\ 4\ часа\ они\ пройдут:\]

\[4x + 4y + 4 = 40\]

\[4 \cdot (x + y) = 36\]

\[x + y = 9.\]

\[До\ середины\ пути\ второй\ \]

\[пешеход\ мог\ дойти\ за\ \frac{20}{y}\ ч,\ \]

\[первый - за\ \frac{20}{x}\ ч.\ Первый\ бы\]

\[\ затратил\ на\ дорогу\ \]

\[на\ 1\ ч\ больше:\]

\[\frac{20}{x} - \frac{20}{y} = 1\]

\[Составим\ систему\ уравнений6\ \]

\[\left\{ \begin{matrix} x + y = 9\ \ \ \ \ \\ \frac{20}{x} - \frac{20}{y} = 1 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 9 - x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{20^{\backslash 9 - x}}{x} - \frac{20^{\backslash x}}{9 - x} = 1^{\backslash x(9 - x)} \\ \end{matrix} \Longrightarrow \right.\ \]

\[\Longrightarrow \left\{ \begin{matrix} y = 9 - x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 180 - 20x - 20x - 9x + x^{2} = 0 \\ \end{matrix} \right.\ \]

\[x^{2} - 49x + 180 = 0\]

\[x_{1} + x_{2} = 49;\ \ \ x_{1} \cdot x_{2} = 180\]

\[x_{1} = 4;\ \ \ \ \ \ x_{2} = 45;\]

\[\left\{ \begin{matrix} x_{1} = 4 \\ y_{1} = 5 \\ \end{matrix} \right.\ \ \ или\ \ \left\{ \begin{matrix} x_{2} = 45\ \ \ \ \\ y_{2} = - 36 \\ \end{matrix} \right.\ ;\ \ \]

\[но\ y > 0 \Longrightarrow не\ подходит.\]

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