Решите уравнение: x(x+1)=24/(x-1)(x+2) (подстановка y=x^2+x).

\[x(x + 1) = \frac{24}{(x - 1)(x + 2)}\]

\[x^{2} + x = \frac{24}{x^{2} + x - 2}\]

\[Пусть\ y = x^{2} + x:\]

\[y = \frac{24}{y - 2}\]

\[ОДЗ:\ \ y \neq 2\]

\[y(y - 2) = 24\]

\[y^{2} - 2y - 24 = 0\]

\[y_{1} + y_{2} = 2\]

\[y_{1} \cdot y_{2} = - 24 \Longrightarrow y_{1} = 6;\ \ y_{2} =\]

\[= - 4\]

\[1)\ x² + x = 6\]

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

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

\[x_{1} \cdot x_{2} = - 6 \Longrightarrow x_{1} = - 3;\ \ x_{2} =\]

\[= 2\]

\[2)\ x² + x = - 4\]

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

\[D = b^{2} - 4ac =\]

\[= 1 - 4 \cdot 1 \cdot 4 < 0 \Longrightarrow\]

\[\Longrightarrow нет\ решения.\]

\[Ответ:x = - 3;\ \ x = 2.\]