Answer
$-\frac{(x-4)^2}{4x} $
Work Step by Step
We apply Method 1 (A51), to treat the numerator and denominator of the complex rational expression separately. So by factoring both the denominator and numerator of each rational function and cancelling the common factors, we have
$$
\frac{\frac{4-x}{4+x}}{\frac{4x}{x^2-16}}=\frac{\frac{4-x}{4+x}}{\frac{4x}{(x+4)(x-4)}}= \frac{4-x}{4+x}\frac{(x+4)(x-4)}{4x}\\
=-\frac{(x-4)^2}{4x}
.$$
