Simplify the following:

$\begin{matrix} \text{(i)} & \sqrt{45} - 3\sqrt{20} + 4\sqrt{5} \\[1.5em] \text{(ii)} & 3\sqrt{3} + 2\sqrt{27} + \dfrac{7}{\sqrt{3}} \\[1.5em] \text{(iii)} & 6\sqrt{5} × 2\sqrt{5} \\[1.5em] \text{(iv)} & 8\sqrt{15} ÷ 2\sqrt{3} \\[1.5em] \text{(v)} & \dfrac{\sqrt{24}}{8} + \dfrac{\sqrt{54}}{9} \\[1.5em] \text{(vi)} & \dfrac{3}{\sqrt{8}} + \dfrac{1}{\sqrt{2}} \\[1.5em] \end{matrix}$

$\text{(i) } \sqrt{45} - 3\sqrt{20} + 4\sqrt{5} \\[1.5em] =\sqrt{3 × 3 × 5} - 3\sqrt{5 × 4} + 4\sqrt{5} \\[1.5em] = 3\sqrt{5} - 3 × 2\sqrt{5} + 4\sqrt5 \\[1.5em] = 3\sqrt{5} - 6\sqrt{5} + 4\sqrt5 \\[1.5em] = \sqrt{5}(3 - 6 + 4) \\[1.5em] = \bold{\sqrt{5}}$

$\text{(ii) } 3\sqrt{3} + 2\sqrt{27} + \dfrac{7}{\sqrt{3}} \\[1.5em] = 3\sqrt{3} + 2\sqrt{3 × 3 × 3} + \dfrac{7}{\sqrt{3}} \\[1.5em] = 3\sqrt{3} + 2 × 3\sqrt{3} + \dfrac{7}{\sqrt{3}} \\[1.5em] = 3\sqrt{3} + 6\sqrt{3} + \dfrac{7}{\sqrt{3}} × \dfrac{\sqrt3}{\sqrt3} \\[1.5em] = \sqrt{3} × (3 + 6 + \dfrac{7}{3}) = \bold{\dfrac{34}{3}{\sqrt{3}}} \\[1.5em]$

$\text{(iii) } 6\sqrt{5} × 2\sqrt{5} \\[1.5em] = 12 × (\sqrt{5} × \sqrt{5}) \\[1.5em] = 12 × (\sqrt{5})^2 \\[1.5em] = 12 × 5 \\[1.5em] = \bold{60} \\[1.5em]$

$\text{(iv) } 8\sqrt{15} ÷ 2\sqrt{3} \\[1.5em] = \dfrac{8\sqrt{15}}{2\sqrt{3}} \\[1.5em] = \dfrac{8\sqrt{3×5}}{2\sqrt{3}} \\[1.5em] = \dfrac{8\sqrt{3}\sqrt{5}}{2\sqrt{3}} \\[1.5em] = \dfrac{8\sqrt{5}}{2} \\[1.5em] = \bold{4\sqrt{5}} \\[1.5em]$

$\text{(v) } \dfrac{\sqrt{24}}{8} + \dfrac{\sqrt{54}}{9} \\[1.5em] = \dfrac{\sqrt{2 × 2 × 6}}{8} + \dfrac{\sqrt{3 × 3 × 6}}{9} \\[1.5em] = \dfrac{2\sqrt{6}}{8} + \dfrac{3\sqrt6}{9}\\[1.5em] = \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt6}{3} \\[1.5em] = \sqrt{6} × {(\dfrac{1}{4} + \dfrac{1}{3})} \\[1.5em] = \sqrt{6} × (\dfrac{3 + 4}{12}) \\[1.5em] = \bold{\dfrac{7}{12}{\sqrt{6}}}$

$\text{(vi) } \dfrac{3}{\sqrt{8}} + \dfrac{1}{\sqrt{2}} \\[1.5em] = \dfrac{3}{\sqrt{2 × 2 × 2}} + \dfrac{1}{\sqrt{2}} \\[1.5em] = \dfrac{3}{2\sqrt{2}} + \dfrac{1}{\sqrt2} \\[1.5em] = \dfrac{1}{\sqrt2} × (\dfrac{3}{2} + 1) \\[1.5em] = \dfrac{1}{\sqrt2} × (\dfrac{3 + 2}{2}) \\[1.5em] = \dfrac{1}{\sqrt2} × \dfrac{5}{2} \\[1.5em] = \dfrac{1}{\sqrt2} × \dfrac{\sqrt{2}}{\sqrt{2}} × \dfrac{5}{2} \\[1.5em] = \bold{\dfrac{5\sqrt{2}}{4}} \\[1.5em]$