\[\boxed{\text{178\ (178).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \frac{x}{x - 2} - \frac{8}{x + 5} = \frac{14}{x^{2} + 3x - 10}\]
\[x^{2} + 3x - 10 = 0\]
\[D = 3^{2} + 4 \cdot 10 = 9 + 40 = 49\]
\[x_{1} = \frac{- 3 - 7}{2} = - 5;\ \ x_{2} =\]
\[= \frac{- 3 + 7}{2} = 2\]
\[x^{2} + 3x - 10 = (x + 5)(x - 2);\]
\[\Longrightarrow \frac{x}{x - 2} - \frac{8}{x + 5} =\]
\[= \frac{14}{(x + 5)(x - 2)};\]
\[\frac{x^{2} + 5x - 8x + 16}{(x - 2)(x + 5)} -\]
\[- \frac{14}{(x + 5)(x - 2)} = 0\]
\[ОДЗ:\ \ \ x \neq - 5;\ \ x \neq 2.\]
\[x^{2} + 5x - 8x + 16 - 14 = 0\]
\[x^{2} - 3x + 2 = 0\]
\[D = 3^{2} - 4 \cdot 2 = 9 - 8 = 1\]
\[x_{1} = \frac{3 + 1}{2} = 2;\ x_{2} = \frac{3 - 1}{2} = 1;\ \ \]
\[x = 2 \Longrightarrow не\ удовлетворяет\ ОДЗ.\]
\[Ответ:\ \ x = 1.\]
\[\textbf{б)}\ \frac{y}{2y - 3} + \frac{1}{y + 7} +\]
\[+ \frac{17}{2y^{2} + 11y - 21} = 0\]
\[2y^{2} + 11y - 21 = 0\]
\[D = 11^{2} + 4 \cdot 2 \cdot 21 = 289\]
\[y_{1} = \frac{- 11 - 17}{4} = - 7;\ \ \ \ y_{2} =\]
\[= \frac{- 11 + 17}{4} = \frac{6}{4} = \frac{3}{2}.\]
\[2y^{2} + 11y - 21 =\]
\[= 2 \cdot (y + 7)\left( y - \frac{3}{2} \right) =\]
\[= (y + 7)(2y - 3)\]
\[\Longrightarrow \frac{y}{2y - 3} + \frac{1}{y + 7} +\]
\[+ \frac{17}{(y + 7)(2y - 3)} = 0\]
\[y^{2} + 7y + 2y - 3 + 17 = 0\]
\[ОДЗ:\ \ x \neq - 7;\ \ \ x \neq \frac{3}{2}.\]
\[y^{2} + 9y + 14 = 0\]
\[D = 9^{2} - 4 \cdot 14 =\]
\[= 81 - 56 = 25\]
\[y_{1} = \frac{- 9 - 5}{2} = - 7;\ \ \ \ y_{2} =\]
\[= \frac{- 9 + 5}{2} = - 2.\ \]
\[y = - 7 \Longrightarrow не\ удовлетворяет\ ОДЗ.\]
\[Ответ:y = - 2.\]
\[\boxed{\text{178.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[y = \frac{x + 2}{x - 1} \rightarrow а);\]
\[y = \frac{- x - 2}{x - 1} \rightarrow б).\]