Решите уравнение: 14/(x^3+x^2-9x-9)-1/(x+3)=7/(x-3)(x+1).

\[\frac{14}{x^{3} + x^{2} - 9x - 9} - \frac{1}{x + 3} =\]

\[= \frac{7}{(x - 3)(x + 1)}\]

\[\frac{4}{x^{2}(x + 1) - 9 \cdot (x + 1)} - \frac{1}{x + 3} =\]

\[= \frac{7}{(x - 3)(x + 1)}\]

\[\frac{14}{(x + 1)(x - 3)(x + 3)} - \frac{1}{x + 3} =\]

\[= \frac{7}{(x - 3)(x + 1)}\]

\[ОДЗ:\ \ x - 3 \neq 0 \Longrightarrow x \neq 3\]

\[\ \ \ \ \ \ \ \ x + 1 \neq 0 \Longrightarrow x \neq - 1\]

\[\ \ \ \ \ \ x + 3 \neq 0 \Longrightarrow x \neq - 3\]

\[\frac{14 - (x + 1)(x - 3) - 7 \cdot (x + 3)}{(x + 1)(x - 3)(x + 3)} = 0\]

\[14 - x^{2} + 3x - x + 3 - 7x - 21 = 0\]

\[- x^{2} - 5x - 4 = 0\]

\[x^{2} + 5x + 4 = 0\]

\[x_{1} + x_{2} = - 5\]

\[x_{1} \cdot x_{2} = 4\]

\[x_{1} = - 1\ (не\ подходит);\ \ \ \]

\[x_{2} = - 4\]

\[Ответ:x = - 4.\]