Answer
$\frac{3x}{x-4}$, $x\ne0, \pm4$
Work Step by Step
Step 1. To remove the fractions, we can multiply both the numerator and the denominator with a common factor $x^2$.
Step 2. We have $\frac{3+\frac{12}{x}}{1-\frac{16}{x^2}}\cdot \frac{x^2}{x^2}=\frac{3x^2+12x}{x^2-16}=\frac{3x(x+4)}{(x+4)(x-4)}$
Step 3. Provided that $x\ne0, \pm4$, we simplify the above result to get $\frac{3x}{x-4}$
Step 4. We conclude the result as $\frac{3x}{x-4}$, with $x\ne0, \pm4$
