Answer
$\dfrac{x+5}{x+4}$
Work Step by Step
The factored form of the given expression, $
\dfrac{x^2-9}{x^2+7x+12}\div\dfrac{x-3}{x+5}
$, is
\begin{align*}
&
\dfrac{(x+3)(x-3)}{x^2+7x+12}\div\dfrac{x-3}{x+5}
&(\text{use }x^2-y^2=(x+y)(x-y)
\\\\&=
\dfrac{(x+3)(x-3)}{(x+3)(x+4)}\div\dfrac{x-3}{x+5}
&(\text{use factoring of trinomials})
.\end{align*}
Multiplying by the reciprocal of the divisor, the expression above is equivalent to
\begin{align*}
&
\dfrac{(x+3)(x-3)}{(x+3)(x+4)}\cdot\dfrac{x+5}{x-3}
.\end{align*}
Cancelling the common terms between the numerator and the denominator, the expression above is equivalent to
\begin{align*}\require{cancel}
&
\dfrac{(\cancel{x+3})(\cancel{x-3})}{(\cancel{x+3})(x+4)}\cdot\dfrac{x+5}{\cancel{x-3}}
\\\\&=
\dfrac{x+5}{x+4}
.\end{align*}
Hence, the expression $
\dfrac{x^2-9}{x^2+7x+12}\div\dfrac{x-3}{x+5}
$ simplifies to $
\dfrac{x+5}{x+4}
$.
