ГДЗ по алгебре 9 класс Макарычев Задание 178

\[\boxed{\text{178\ (178).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{x}{x - 2} - \frac{8}{x + 5} = \frac{14}{x^{2} + 3x - 10}\]

\[x^{2} + 3x - 10 = 0\]

\[D = 3^{2} + 4 \cdot 10 = 9 + 40 = 49\]

\[x_{1} = \frac{- 3 - 7}{2} = - 5;\ \ x_{2} =\]

\[= \frac{- 3 + 7}{2} = 2\]

\[x^{2} + 3x - 10 = (x + 5)(x - 2);\]

\[\Longrightarrow \frac{x}{x - 2} - \frac{8}{x + 5} =\]

\[= \frac{14}{(x + 5)(x - 2)};\]

\[\frac{x^{2} + 5x - 8x + 16}{(x - 2)(x + 5)} -\]

\[- \frac{14}{(x + 5)(x - 2)} = 0\]

\[ОДЗ:\ \ \ x \neq - 5;\ \ x \neq 2.\]

\[x^{2} + 5x - 8x + 16 - 14 = 0\]

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

\[D = 3^{2} - 4 \cdot 2 = 9 - 8 = 1\]

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

\[x = 2 \Longrightarrow не\ удовлетворяет\ ОДЗ.\]

\[Ответ:\ \ x = 1.\]

\[\textbf{б)}\ \frac{y}{2y - 3} + \frac{1}{y + 7} +\]

\[+ \frac{17}{2y^{2} + 11y - 21} = 0\]

\[2y^{2} + 11y - 21 = 0\]

\[D = 11^{2} + 4 \cdot 2 \cdot 21 = 289\]

\[y_{1} = \frac{- 11 - 17}{4} = - 7;\ \ \ \ y_{2} =\]

\[= \frac{- 11 + 17}{4} = \frac{6}{4} = \frac{3}{2}.\]

\[2y^{2} + 11y - 21 =\]

\[= 2 \cdot (y + 7)\left( y - \frac{3}{2} \right) =\]

\[= (y + 7)(2y - 3)\]

\[\Longrightarrow \frac{y}{2y - 3} + \frac{1}{y + 7} +\]

\[+ \frac{17}{(y + 7)(2y - 3)} = 0\]

\[y^{2} + 7y + 2y - 3 + 17 = 0\]

\[ОДЗ:\ \ x \neq - 7;\ \ \ x \neq \frac{3}{2}.\]

\[y^{2} + 9y + 14 = 0\]

\[D = 9^{2} - 4 \cdot 14 =\]

\[= 81 - 56 = 25\]

\[y_{1} = \frac{- 9 - 5}{2} = - 7;\ \ \ \ y_{2} =\]

\[= \frac{- 9 + 5}{2} = - 2.\ \]

\[y = - 7 \Longrightarrow не\ удовлетворяет\ ОДЗ.\]

\[Ответ:y = - 2.\]