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МатематикаГДЗ по алгебре 8 класс Макарычев ФГОС Задание 635
Задание 635
\[\boxed{\text{635\ (635).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[Пусть\ \text{x\ }\frac{км}{ч} - скорость\ \]
\[мотоциклиста\ на\ первой\ \]
\[половине\ пути,\ тогда\ \]
\[(x - 20)\ \frac{км}{ч} - скорость\ \]
\[после\ снижения.\]
\[Известно,\ что\ средняя\ \]
\[скорость\ на\ всем\ пути\ \]
\[составила\ 37,5\ \frac{км}{ч}.\]
\[Составим\ уравнение:\]
\[\frac{2}{\frac{1}{x} + \frac{1}{x - 20}} = 37,5\]
\[\frac{2}{\frac{x - 20 + x}{x^{2} - 20x}} = 37,5\]
\[\frac{2 \cdot \left( x^{2} - 20x \right)}{2x - 20} = 37,5\]
\[\frac{2 \cdot \left( x^{2} - 20x \right)}{2 \cdot (x - 10)} = 37,5\]
\[x^{2} - 20x = 37,5 \cdot (x - 10)\]
\[x^{2} - 20x = 37,5x - 375\]
\[x^{2} - 57,5x + 375 = 0\ \ \ \ \ \ | \cdot 2\]
\[2x^{2} - 115x + 750 = 0\]
\[D = 13\ 225 - 600 = 7225\]
\[x_{2} = \frac{115 + 85}{4} = 50\ \left( \frac{км}{ч} \right) -\]
\[скорость\ мотоциклиста\ \]
\[на\ первой\ половине\ пути.\]
\[Ответ:50\frac{км}{ч}.\]
\[\boxed{\text{635.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
\[\textbf{а)}\ y = \frac{2x - 1}{x + 6}\]
\[y = 5:\]
\[5 = \frac{2x - 1}{x + 6}\]
\[5 \cdot (x + 6) = 2x - 1\]
\[5x + 30 = 2x - 1\]
\[5x - 2x = - 1 - 30\]
\[3x = - 31\]
\[x = - \frac{31}{3} = - 10\frac{1}{3}.\]
\[Ответ:при\ x = - 10\frac{1}{3}.\]
\[y = - 3:\]
\[- 3 = \frac{2x - 1}{x + 6}\]
\[- 3 \cdot (x + 6) = 2x - 1\]
\[- 3x - 18 = 2x - 1\]
\[- 3x - 2x = - 1 + 18\]
\[- 5x = 17\]
\[x = - \frac{17}{5} = - 3,4\]
\[Ответ:при\ x = - 3,4.\]
\[y = 0:\]
\[0 = \frac{2x - 1}{x + 6}\]
\[0 = 2x - 1\]
\[2x = 1\]
\[x = 0,5\]
\[Ответ:при\ x = 0,5.\]
\[y = 2:\]
\[2 = \frac{2x - 1}{x + 6}\]
\[2 \cdot (x + 6) = 2x - 1\]
\[2x + 12 = 2x - 1\]
\[2x - 2x = - 1 - 12\]
\[0x = - 13\]
\[Ответ:нет\ корней.\]
\[\textbf{б)}\ y = \frac{x^{2} + x - 2}{x + 3}\]
\[y = - 10:\]
\[- 10 = \frac{x^{2} + x - 2}{x + 3}\]
\[- 10 \cdot (x + 3) = x^{2} + x - 2\]
\[- 10x - 30 = x^{2} + x - 2\]
\[x^{2} + 11x + 28 = 0\]
\[D = 121 - 112 = 9\]
\[x_{1} = \frac{- 11 - 3}{2} = - 7;\ \ \ \]
\[x_{2} = \frac{- 11 + 3}{2} = - 4\]
\[Ответ:при\ x = - 7;\ \ x = - 4.\]
\[y = 0:\]
\[0 = \frac{x^{2} + x - 2}{x + 3}\]
\[x^{2} + x - 2 = 0\]
\[x_{1} + x_{2} = - 1;\ \ \ x_{1} \cdot x_{2} = - 2\]
\[x_{1} = - 2;\ \ \ x_{1} = 1\]
\[Ответ:при\ x = - 2;\ \ x = 1.\]
\[y = - 5:\]
\[- 5 = \frac{x^{2} + x - 2}{x + 3}\]
\[- 5 \cdot (x + 3) = x^{2} + x - 2\]
\[- 5x - 15 = x^{2} + x - 2\]
\[x^{2} + 6x + 13 = 0\]
\[D = 36 - 52 = - 16 < 0\]
\[Ответ:корней\ нет.\]
