Решебник по алгебре 9 класс Мерзляк Задание 995

\[\boxed{\mathbf{995\ (995).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[2ч\ 15\ мин = 2\frac{15}{60} = 2\frac{1}{4} = \frac{9}{4}\ ч.\]

\[\frac{9}{4}\ ч - 1\ ч = \frac{5}{4}\ (ч) - грузовик\]

\[\ ехал\ дольше.\]

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

\[\ автобуса,\ \text{y\ }\frac{км}{ч} - грузовика.\]

\[Составим\ систему:\]

\[\left\{ \begin{matrix} \frac{180}{y} - \frac{180}{x} = \frac{5}{4} \\ x + \frac{9}{4}y = 180\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 180 \cdot \left( \frac{1}{y} - \frac{1}{x} \right) = \frac{5}{4} \\ x = 180 - \frac{9}{4}\text{y\ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} \frac{1}{y} - \frac{1}{x} = \frac{1}{144} \\ x = 180 - \frac{9}{4}y \\ \end{matrix} \right.\ \]

\[\frac{1}{y} - \frac{1}{180 - \frac{9}{4}y} = \frac{1}{144}\]

\[\frac{180 - \frac{9}{4}y - y}{\left( 180 - \frac{9}{4}y \right) \cdot y} = \frac{1}{144}\]

\[\frac{180 - \frac{13}{4}y}{y \cdot \left( 180 - \frac{9}{4}y \right)} = \frac{1}{144}\]

\[144 \cdot \left( 180 - \frac{13}{4}y \right) =\]

\[= y\left( 180 - \frac{9}{4}y \right)\]

\[25\ 920 - 468y - 180y +\]

\[+ \frac{9}{4}y^{2} = 0\ \ \ | \cdot \frac{4}{9}\]

\[y^{2} - 288y + 11520 = 0\]

\[y_{1} + y_{2} = 288,\ \ \]

\[y_{1}y_{2} = 11\ 520\]

\[y = 240 - не\ удовлетворяет.\]

\[y = 48\]

\[\left\{ \begin{matrix} y = 48\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x = 180 - \frac{9}{4} \cdot 48 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = 48 \\ x = 72 \\ \end{matrix} \right.\ \]

\[Ответ:72\frac{км}{ч},\ 48\ \frac{км}{ч}\text{.\ }\]