Решите уравнение: (x^2-16)/(x^3+4x^2-3x-12)=0.

\[\frac{x^{2} - 16}{x^{3} + 4x^{2} - 3x - 12} = 0\]

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

\[x - 4 = 0\]

\[x = 4\]

\[Ответ:4.\]