Divide the sum of $\dfrac{65}{12} \text{ and } \dfrac{8}{3}$ by their difference.

[Hint: Here we have to find $\left(\dfrac{65}{12} + \dfrac{8}{3}\right) \div \left(\dfrac{65}{12} - \dfrac{8}{3}\right)$]

First find the sum and the difference of the given rational numbers.

LCM of 12 and 3 = 12

Finding the sum,

$\dfrac{65}{12} + \dfrac{8}{3} = \dfrac{65 + 8 \times 4}{12} \\[1em] = \dfrac{65 + 32}{12} \\[1em] = \dfrac{97}{12}$

Finding the difference,

$\dfrac{65}{12} - \dfrac{8}{3} = \dfrac{65 - 8 \times 4}{12} \\[1em] = \dfrac{65 - 32}{12} \\[1em] = \dfrac{33}{12} = \dfrac{11}{4}$

Dividing the sum by the difference,

$\dfrac{97}{12} \div \dfrac{11}{4} = \dfrac{97}{12} \times \dfrac{4}{11} \\[1em] = \dfrac{97 \times 4}{12 \times 11} \\[1em] = \dfrac{388}{132} = \dfrac{97}{33} \\[1em] = 2\dfrac{31}{33}$

Hence, the required quotient is $\mathbf{2\dfrac{31}{33}}$.