Simplify :

$1\dfrac{1}{3} ÷ \dfrac{3}{7}$ of $2\dfrac{5}{8} + 1\dfrac{1}{9}$

Solving,

$\Rightarrow 1\dfrac{1}{3} ÷ \dfrac{3}{7} \text{ of } 2\dfrac{5}{8} + 1\dfrac{1}{9}\\[1em] \Rightarrow \dfrac{4}{3} ÷ \Big(\dfrac{3}{7} \times \dfrac{21}{8}\Big) + \dfrac{10}{9}\\[1em] \Rightarrow \dfrac{4}{3} ÷ \dfrac{3 \times 21}{7 \times 8} + \dfrac{10}{9}\\[1em] \Rightarrow \dfrac{4}{3} ÷ \dfrac{63}{56} + \dfrac{10}{9}\\[1em] \Rightarrow \dfrac{4}{3} ÷ \dfrac{9}{8} + \dfrac{10}{9}\\[1em] \Rightarrow \dfrac{4}{3} \times \dfrac{8}{9} + \dfrac{10}{9}\\[1em] \Rightarrow \dfrac{4 \times 8}{3 \times 9} + \dfrac{10}{9}\\[1em] \Rightarrow \dfrac{32}{27} + \dfrac{10}{9}\\[1em] \Rightarrow \dfrac{32}{27} + \dfrac{10 \times 3}{9 \times 3}\\[1em] \Rightarrow \dfrac{32}{27} + \dfrac{30}{27}\\[1em] \Rightarrow \dfrac{32 + 30}{27}\\[1em] \Rightarrow \dfrac{62}{27}\\[1em] \Rightarrow 2\dfrac{8}{27}$

Hence, $1\dfrac{1}{3} ÷ \dfrac{3}{7}$ of $2\dfrac{5}{8} + 1\dfrac{1}{9} = 2\dfrac{8}{27}$.