ГДЗ по алгебре 9 класс Макарычев Задание 372

\[\boxed{\text{372\ (372).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left( \frac{x + 1}{x - 2} \right)^{2} - 16 \cdot \left( \frac{x - 2}{x + 1} \right)^{2} = 15\]

\[Пусть\ \ t = \left( \frac{x + 1}{x - 2} \right)^{2}:\]

\[\ t - \frac{16}{t} = 15\]

\[t^{2} - 15t - 16 = 0\]

\[По\ теореме\ Виета:\]

\[t_{1} + t_{2} = 15;\ \ \ t_{1} \cdot t_{2} = - 16\]

\[t_{1} = 16;\ \ \ \ t_{2} = - 1.\]

\[\left( \frac{x + 1}{x - 2} \right)^{2} = 16\ \ \]

\[\frac{x + 1}{x - 2} = \pm 4\]

\[1)\ \ \ \]

\[\frac{x + 1}{x - 2} = 4\ \]

\[x + 1 = 4x - 8\]

\[3x = 9\ \ \]

\[x = 3.\]

\[2)\ \frac{x + 1}{x - 2} = - 4\]

\[x + 1 = - 4x - 8\]

\[5x = 7\]

\[x = 1,4.\]

\[\left( \frac{x + 1}{x - 2} \right)^{2} = - 1 \Longrightarrow корней\ нет.\]

\[Ответ:x = 1,4;\ \ x = 3.\]

\[\textbf{б)}\ \left( \frac{x + 3}{x - 5} \right)^{2} - 9 \cdot \left( \frac{x - 5}{x + 3} \right)^{2} = 8;\]

\[Пусть\ t = \left( \frac{x + 3}{x - 5} \right)^{2}:\ \]

\[t - \frac{9}{t} = 8\]

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

\[D_{1} = 16 + 9 = 25\]

\[t_{1} = 4 + 5 = 9;\ \ \]

\[\ t_{2} = 4 - 5 = - 1.\]

\[\left( \frac{x + 3}{x - 5} \right)^{2} = 9\ \ \]

\[\frac{x + 3}{x - 5} = \pm 3.\]

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

\[x + 3 = 3x - 15\]

\[2x = 18\]

\[x = 9.\]

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

\[x + 3 = - 3x + 15\]

\[4x = 12\]

\[x = 3.\]

\[2)\ \ \left( \frac{x + 3}{x - 5} \right)^{2} = - 1 \Longrightarrow\]

\[\Longrightarrow корней\ нет.\]

\[Ответ:x = 3;x = 9.\]