$\frac{1}{x^2- 9} + \frac{1}{x+3} =($ $)/($ $)$.
$\frac{1}{x^2- 9} - \frac{1}{x+3} =($ $)/($ $)$.
$\frac{1}{x^2- 9} \times \frac{1}{x+3} =($ $)/($ $)$.
$\frac{1}{x^2- 9} \div \frac{1}{x+3} =($ $)/($ $)$.
Hint: The main difference between this and the preceding problems is that the second denominator is a factor of the first denominator.