\[\frac{3x - 9}{x - 1} + \frac{x + 6}{x + 1} = 3\]
\[ОДЗ:\ \ x \neq 1\]
\[\ \ \ \ \ \ \ \ \ \ \ x \neq - 1\]
\[4x^{2} - x - 15 = 3 \cdot \left( x^{2} - 1 \right)\]
\[4x^{2} - x - 15 - 3x^{2} + 3 = 0\]
\[x^{2} - x - 12 = 0\]
\[x_{1} + x_{2} = 1\]
\[x_{1} \cdot x_{2} = - 12 \Longrightarrow x_{1} = 4;\ \ \]
\[x_{2} = - 3\]
\[Ответ:\ x = 4;\ \ x = - 3.\]