\[\boxed{\mathbf{814\ (814).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[Пусть\ x\ \frac{км}{ч} - скорость\ \]
\[автомобиля,\ а\ (x - 20)\ \frac{км}{ч} -\]
\[скорость\ катера.\ Тогда\ время\ \]
\[в\ пути\ на\ катере\ \]
\[\frac{5}{8} \cdot 160\ :(x - 20)\ ч,\ а\ на\ \]
\[автомобиле - \frac{3}{8} \cdot 160\ :x\ \ ч.\ \]
\[По\ условию\ известно,\ что\ на\ \]
\[автомобиле\ турист\ ехал\ \]
\[на\ 1,5\ часа\ меньше.\]
\[Составляем\ уравнение:\ \]
\[\frac{100}{x - 20} - \frac{60}{x} - 1,5 = 0;\ \ \ \ \ x \neq 0;\ \ \ \ \]
\[x \neq 20\]
\[- 1,5x^{2} + 70x + 1200 = 0\ \ \ \ \ \]
\[D = 4900 + 7200 = 12\ 100\]
\[x = \frac{- 70 - 110}{- 3} = 60\ \left( \frac{км}{ч} \right) -\]
\[скорость\ автомобиля.\]
\[x = \frac{- 70 + 110}{- 3} = - \frac{40}{3} \Longrightarrow не\ \]
\[удовлетворяет\ условию.\]
\[60 - 20 = 40\ \left( \frac{км}{ч} \right) -\]
\[скорость\ катера.\]
\[Ответ:60\ \frac{км}{ч},\ 40\ \frac{км}{ч}.\]
\[\boxed{\mathbf{81}\mathbf{4}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ \frac{x^{2} - 6x}{x - 4} = 4\]
\[\frac{x^{2} - 6x - 4x + 16}{x - 4} = 0;\ \ \ \ \ x \neq 4\]
\[x^{2} - 10x + 16 = 0\]
\[x_{1} + x_{2} = 10,\ \ x_{1} = 8\]
\[x_{1} \cdot x_{2} = 16,\ \ x_{2} = 2\]
\[Ответ:x = 8;x = 2.\]
\[2)\ \frac{5x + 18}{x - 2} = x\]
\[\frac{5x + 18}{x - 2} - x = 0\]
\[\frac{5x + 18 - x^{2} + 2x}{x - 2} = 0;\ \ \ \ \ x \neq 2\]
\[- x^{2} + 7x + 18 = 0\]
\[x^{2} - 7x - 18 = 0\]
\[x_{1} + x_{2} = 7,\ \ x_{1} = 9\]
\[x_{1} \cdot x_{2} = - 18,\ \ x_{2} = - 2\]
\[Ответ:x = 9;\ x = - 2.\]
\[3)\ x + 1 = \frac{6}{x}\ \ \ \ \ \ \ \ | \cdot x\ \ \ \ \ \]
\[\ \ \ \ \ \ x \neq 0\]
\[x^{2} + x - 6 = 0\]
\[x_{1} + x_{2} = - 1,\ \ x_{1} = - 3\]
\[x_{1} \cdot x_{2} = - 6,\ \ x_{2} = 2\]
\[Ответ:x = 2;\ x = - 3.\]
\[4)\ 5 - \frac{8}{x^{2}} = \frac{18}{x}\]
\[5 - \frac{8}{x^{2}} - \frac{18}{x} = 0\]
\[\frac{5x² - 8 - 18x}{x²} = 0;\ \ \ \ \ \ \ \ x \neq 0\]
\[5x^{2} - 18x - 8 = 0\]
\[D = 324 + 160 = 484,\ \ \]
\[x_{1,2} = \frac{18 \pm 22}{10}\]
\[x_{1} = 4;\ \ \ \ \ \ x_{2} = - 0,4\]
\[Ответ:\ x = - 0,4;x = 4.\]