Найдите корни уравнения: 2/(x-4)-5/(x+3)=11/(x^2-x-12)+1.

\[\frac{2}{x - 4} - \frac{5}{x + 3} = \frac{11}{x^{2} - x - 12} + 1\]

\[x^{2} - x - 12 = (x + 3)(x - 4)\]

\[x_{1} + x_{2} = 1;\ \ \ x_{1} \cdot x_{2} = - 12\]

\[x_{1} = 4;\ \ x_{2} = - 3.\]

\[\frac{- x^{2} - 2x + 27}{(x + 3)(x - 4)} = 0;\ \ \ \ \ \ \]

\[\ x \neq - 3;\ \ x \neq 4\]

\[x^{2} + 2x - 27 = 0\]

\[D = 1 + 27 = 28 = \left( 2\sqrt{7} \right)^{2}\]

\[x_{1} = - 1 + 2\sqrt{7};\ \ \ \]

\[x_{2} = - 1 - 2\sqrt{7}.\]

\[Ответ:x = - 1 \pm 2\sqrt{7}.\]