Answer
$\dfrac{x+3}{4x^{2}-9}\div\dfrac{x^{2}+7x+12}{2x^{2}+7x-15}=\dfrac{x+5}{(2x+3)(x+4)}$
Work Step by Step
$\dfrac{x+3}{4x^{2}-9}\div\dfrac{x^{2}+7x+12}{2x^{2}+7x-15}$
Factor the denominator of the first fraction and the numerator and the denominator of the second fraction:
$\dfrac{x+3}{(2x+3)(2x-3)}\div\dfrac{(x+4)(x+3)}{(2x-3)(x+5)}=...$
Evaluate the division and simplify:
$...=\dfrac{(x+3)(2x-3)(x+5)}{(2x+3)(2x-3)(x+4)(x+3)}=\dfrac{x+5}{(2x+3)(x+4)}$
