Решебник по алгебре 8 класс Макарычев ФГОС Задание 634

 

\[\boxed{\text{634\ (634).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[велосипедиста\ до\ поселка,\ \]

\[тогда\ (x + 5)\ \frac{км}{ч} - скорость\ \]

\[велосипедиста\ на\ обратном\ \]

\[пути.\]

\[Известно,\ что\ средняя\ \]

\[скорость\ на\ всем\ пути\ \]

\[следования\ 12\ \frac{км}{ч}.\]

\[Составим\ уравнение:\]

\[\frac{2}{\frac{1}{x} + \frac{1}{x + 5}} = 12\]

\[\frac{2}{\frac{x + 5 + x}{x^{2} + 5x}} = 12\]

\[2 \cdot \frac{x^{2} + 5x}{2x + 5} = 12\]

\[\frac{x^{2} + 5x}{2x + 5} = 6\]

\[x^{2} + 5x = 6 \cdot (2x + 5)\]

\[x^{2} + 5x = 12x + 30\]

\[x^{2} - 7x - 30 = 0\]

\[D = 49 + 120 = 169\]

\[x_{2} = \frac{7 + 13}{2} = 10\ \left( \frac{км}{ч} \right) -\]

\[скорость\ велосипедиста\ \]

\[до\ поселка.\]

\[Ответ:10\ \frac{км}{ч}\text{.\ \ }\]

\[\boxed{\text{634.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ \frac{3x + 1}{x + 2} - \frac{x - 1}{x - 2} = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \]

\[x + 2 \neq 0\ \ \ и\ \ \ x - 2 \neq 0\]

\[\ \ \ \ \ \ \ \ x \neq - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq 2\]

\[3x^{2} - 6x + x - 2 - x^{2} - 2x +\]

\[+ x + 2 = x^{2} - 4\]

\[x^{2} - 6x + 4 = 0\]

\[D = 36 - 16 = 20\]

\[x_{1,2} = \frac{6 \pm \sqrt{20}}{2} =\]

\[= \frac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5}\]

\[Ответ:x = \left\{ 3 - \sqrt{5};3 + \sqrt{5} \right\}.\]

\[\textbf{б)}\frac{2y - 2}{y + 3} + \frac{y + 3}{y - 3} = 5\]

\[y + 3 \neq 0\ \ \ и\ \ \ y - 3 \neq 0\]

\[\ \ \ \ \ \ \ \ y \neq - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ y \neq 3\]

\[2y^{2} - 6y - 2y + 6 + y^{2} +\]

\[+ 6y + 9 = 5y^{2} - 45\]

\[- 2y^{2} - 2y + 60 = 0\ \ \ \ |\ :( - 2)\]

\[y^{2} + y - 30 = 0\]

\[D = 1 + 120 = 121\]

\[y_{1,2} = \frac{- 1 \pm 11}{2} = 5;\ - 6.\]

\[Ответ:y = \left\{ - 6;5 \right\}.\]

\[\textbf{в)}\ \frac{4}{9y^{2} - 1} - \frac{4}{3y + 1} = \frac{5}{1 - 3y}\]

\[9y^{2} - 1 \neq 0,\ \ 9y^{2} \neq 1,\]

\[\text{\ \ }y^{2} \neq \frac{1}{9},\ \ y \neq \pm \frac{1}{3}\]

\[3y + 1 \neq 0,\ \ 3y \neq - 1,\ \]

\[\ y \neq - \frac{1}{3}\]

\[1 - 3y \neq 0,\ \ - 3y \neq - 1,\]

\[\ \ y \neq \frac{1}{3}\]

\[\frac{4}{(3y - 1)(3y + 1)} -\]

\[- \frac{4}{3y + 1} = \frac{- 5}{3y - 1}\]

\[4 - 4 \cdot (3y - 1) = - 5 \cdot (3y + 1)\]

\[4 - 12y + 4 = - 15y - 5\]

\[3y = - 13\]

\[y = - \frac{13}{3}\]

\[Ответ:y = - 4\frac{1}{3}.\]

\[\textbf{г)}\frac{4}{x + 3} - \frac{5}{3 - x} = \frac{1}{x - 3} - 1\]

\[x + 3 \neq 0,\ \ x \neq - 3\]

\[x - 3 \neq 0,\ \ x \neq 3\]

\[\frac{4}{x + 3} + \frac{5}{x - 3} = \frac{1 - x + 3}{x - 3}\]

\[4 \cdot (x - 3) + 5 \cdot (x + 3) =\]

\[= (4 - x)(x + 3)\]

\[4x - 12 + 5x + 15 =\]

\[= 4x + 12 - x^{2} - 3x\]

\[x^{2} + 8x - 9 = 0\]

\[D = 64 + 36 = 100\]

\[x_{1,2} = \frac{- 8 \pm 10}{2} = 1;\ - 9.\]

\[Ответ:x = \left\{ - 9;1 \right\}.\]

\[\textbf{д)}\frac{3}{x} + \frac{4}{x - 1} = \frac{5 - x}{x^{2} - x}\]

\[x \neq 0\]

\[x - 1 \neq 0,\ \ x \neq 1\]

\[\frac{3}{x} + \frac{4}{x - 1} = \frac{5 - x}{x(x - 1)}\]

\[3x - 3 + 4x = 5 - x\]

\[8x = 8\]

\[x = 1,\ так\ как\ по\ ОДЗ\ x \neq 1,\ \]

\[то\ корней\ нет.\]

\[Ответ:корней\ нет.\]

\[\textbf{е)}\frac{3y - 2}{y} - \frac{1}{y - 2} = \frac{3y + 4}{y^{2} - 2y}\]

\[y \neq 0\]

\[y - 2 \neq 0,\ \ y \neq 2\]

\[\frac{3y - 2}{y} - \frac{1}{y - 2} = \frac{3y + 4}{y(y - 2)}\]

\[(3y - 2)(y - 2) - y = 3y + 4\]

\[3y^{2} - 6y - 2y + 4 - y =\]

\[= 3y + 4\]

\[3y^{2} - 12y = 0\]

\[3y(y - 4) = 0\]

\[3y = 0\ \ \ и\ \ \ y - 4 = 0\]

\[\ \ y = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y = 4\]

\[y = 0 - не\ подходит\ \]

\[по\ ОДЗ\ y \neq 0\]

\[Ответ:y = 4.\ \]