Answer
$ \frac{2c-1}{3c(c+2)}, with$ $c\ne-2$ $and$ $c\ne0$
Work Step by Step
Given : $\frac{\frac{c+4}{c^{2}+5c+6}}{\frac{3c^{2}+12c}{2c^{2}+5c-3}}$
This becomes : $\frac{c+4}{(c+3)(c+2)} \times \frac{(2c-1)(c+3)}{3c(c+4)}$
$= \frac{2c-1}{3c(c+2)}$
(After dividing out the common factors (c+3) and (c+4))
