ГДЗ по алгебре 8 класс Макарычев Задание 604

\[\boxed{\text{604\ (604).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ y = \frac{2x - 1}{x + 6}\]

\[y = 5:\]

\[5 = \frac{2x - 1}{x + 6}\]

\[5 \cdot (x + 6) = 2x - 1\]

\[5x + 30 = 2x - 1\]

\[5x - 2x = - 1 - 30\]

\[3x = - 31\]

\[x = - \frac{31}{3} = - 10\frac{1}{3}.\]

\[Ответ:при\ x = - 10\frac{1}{3}.\]

\[y = - 3:\]

\[- 3 = \frac{2x - 1}{x + 6}\]

\[- 3 \cdot (x + 6) = 2x - 1\]

\[- 3x - 18 = 2x - 1\]

\[- 3x - 2x = - 1 + 18\]

\[- 5x = 17\]

\[x = - \frac{17}{5} = - 3,4\]

\[Ответ:при\ x = - 3,4.\]

\[y = 0:\]

\[0 = \frac{2x - 1}{x + 6}\]

\[0 = 2x - 1\]

\[2x = 1\]

\[x = 0,5\]

\[Ответ:при\ x = 0,5.\]

\[y = 2:\]

\[2 = \frac{2x - 1}{x + 6}\]

\[2 \cdot (x + 6) = 2x - 1\]

\[2x + 12 = 2x - 1\]

\[2x - 2x = - 1 - 12\]

\[0x = - 13\]

\[Ответ:нет\ корней.\]

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

\[y = - 10:\]

\[- 10 = \frac{x^{2} + x - 2}{x + 3}\]

\[- 10 \cdot (x + 3) = x^{2} + x - 2\]

\[- 10x - 30 = x^{2} + x - 2\]

\[x^{2} + 11x + 28 = 0\]

\[D = 121 - 112 = 9\]

\[x_{1} = \frac{- 11 - 3}{2} = - 7;\]

\[\text{\ \ \ }x_{2} = \frac{- 11 + 3}{2} = - 4\]

\[Ответ:при\ x = - 7;\ \ x = - 4.\]

\[y = 0:\]

\[0 = \frac{x^{2} + x - 2}{x + 3}\]

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

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

\[x_{1} = - 2;\ \ \ x_{1} = 1\]

\[Ответ:при\ x = - 2;\ \ x = 1.\]

\[y = - 5:\]

\[- 5 = \frac{x^{2} + x - 2}{x + 3}\]

\[- 5 \cdot (x + 3) = x^{2} + x - 2\]

\[- 5x - 15 = x^{2} + x - 2\]

\[x^{2} + 6x + 13 = 0\]

\[D = 36 - 52 = - 16 < 0\]

\[Ответ:корней\ нет.\]