Answer
$\frac{3x}{x-4}; x \ne 0,-4,4;$
Work Step by Step
$\frac{3+\frac{12}{x}}{1-\frac{16}{x^{2}}}$
$= \frac{\frac{3x+12}{x}}{\frac{x^{2}-16}{x^{2}}}; x \ne 0,-4,4;$
$= \frac{\frac{3(x+4)}{x}}{\frac{(x+4)(x-4)}{x^{2}}}; x \ne 0,-4,4;$
$= \frac{3(x+4)}{x} \times \frac{x^{2}}{(x+4)(x-4)}$
$=\frac{3x}{x-4}; x \ne 0,-4,4;$
