The range of the relation $R = \left\{(x, y): y = x + \dfrac{6}{x}\ ;\ \text{where } x, y \in \mathbb{N} \text{ and } x < 6\right\}$ is:

Given relation $R = \left\{(x, y): y = x + \dfrac{6}{x}\ ;\ \text{where } x, y \in \mathbb{N} \text{ and } x < 6\right\}$

Since $x \in \mathbb{N}$ and $x < 6$, the possible values of $x$ are $1, 2, 3, 4, 5$.

Also, $y \in \mathbb{N}$, which means $x + \dfrac{6}{x}$ must be a natural number. This is possible only when $x$ is a divisor of $6$.

Therefore, the valid values of $x$ are $1, 2, 3$.

For $x = 1$, $y = 1 + \dfrac{6}{1} = 7$

For $x = 2$, $y = 2 + \dfrac{6}{2} = 5$

For $x = 3$, $y = 3 + \dfrac{6}{3} = 5$

The relation is $R = \{(1, 7), (2, 5), (3, 5)\}$.

The range of the relation is the set of all second elements of the ordered pairs, which is $\{5, 7\}$.

Answer: $\{5, 7\}$