Answer
$\dfrac{x+3}{x^{2}-9}\div\dfrac{5x+15}{(x-3)^{2}}=\dfrac{x-3}{5(x+3)}$
Work Step by Step
$\dfrac{x+3}{x^{2}-9}\div\dfrac{5x+15}{(x-3)^{2}}$
Factor the denominator of the first fraction and take out common factor $5$ from the numerator of the second fraction:
$\dfrac{x+3}{x^{2}-9}\div\dfrac{5x+15}{(x-3)^{2}}=\dfrac{x+3}{(x-3)(x+3)}\div\dfrac{5(x+3)}{(x-3)^{2}}=...$
Evaluate the division of the two rational expressions and simplify by removing the factors that appear both in the numerator and the denominator of the resulting fraction:
$...=\dfrac{(x+3)(x-3)^{2}}{5(x+3)^{2}(x-3)}=\dfrac{x-3}{5(x+3)}$
