Решите уравнение: (x^2-9)/(x^3-3x^2+4x-12)=0.

\[\frac{x^{2} - 9}{x^{3} - 3x^{2} + 4x - 12} = 0\]

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

\[\frac{x + 3}{x^{2} + 4} = 0\]

\[x + 3 = 0\]

\[x = - 3\]

\[Ответ:\ - 3.\]