Answer
$\dfrac{(x+3)^2}{(x-3)^2} $
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
$$
\frac{\frac{x^2+7x+12}{x^2-7x+12}}{\frac{x^2+x-12}{x^2-x-12}}=\frac{\frac{(x+3)(x+4)}{(x-3)(x-4)}}{\frac{(x-3)(x+4)}{(x+3)(x-4)}}=\frac{(x+3)(x+4)}{(x-3)(x-4)}\frac{(x+3)(x-4)}{(x-3)(x+4)}\\
=\frac{(x+3)^2}{(x-3)^2}
.$$
