To remove the radical $\sqrt{2}$ from the denominator, all we need to do is multiply both the numerator and denominator by $\sqrt{2}$ and simplify:

\begin{align} \dfrac {1} {\sqrt{2}} &= \\[5pt] \dfrac {1} {\sqrt{2}} \cdot \dfrac {\sqrt{2}} {\sqrt{2}} &= \\[5pt] \dfrac {1\cdot \sqrt{2}} {( \sqrt{2} )^2} &= \\[5pt] \dfrac {\sqrt{2}} {2} \end{align}

To remove the radical $\sqrt{2}$ from the denominator, all we need to do is multiply both the numerator and denominator by $\sqrt{2}$ and simplify:

\begin{align} \dfrac {5} {\sqrt{2}} &= \\[5pt] \dfrac {5} {\sqrt{2}} \cdot \dfrac {\sqrt{2}} {\sqrt{2}} &= \\[5pt] \dfrac {5\sqrt{2}} {( \sqrt{2} )^2} &= \\[5pt] \dfrac {5\sqrt{2}} {2} \end{align}

To remove the radical $\sqrt{2}$ from the denominator, all we need to do is multiply both the numerator and denominator by $\sqrt{2}$ and simplify:

\begin{align} \dfrac {4} {\sqrt{2}} &= \\[5pt] \dfrac {4} {\sqrt{2}} \cdot \dfrac {\sqrt{2}} {\sqrt{2}} &= \\[5pt] \dfrac {4\sqrt{2}} {( \sqrt{2} )^2} &= \\[5pt] \dfrac {4\sqrt{2}} {2} &= \\[5pt] 2\sqrt{2} & \end{align}

To remove the radical $\sqrt{3}$ from the denominator, all we need to do is multiply both the numerator and denominator by $\sqrt{3}$ and simplify:

\begin{align} \dfrac {9} {2\sqrt{3}} &= \\[5pt] \dfrac {9} {2\sqrt{3}} \cdot \dfrac {\sqrt{3}} {\sqrt{3}} &= \\[5pt] \dfrac {9\sqrt{3}} {2\cdot( \sqrt{3} )^2} &= \\[5pt] \dfrac {9\sqrt{3}} {2\cdot 3} &= \\[5pt] \dfrac {3\sqrt{3}} {2} & \end{align}

To remove the radical $\sqrt{5}$ from the denominator, all we need to do is multiply both the numerator and denominator by $\sqrt{5}$ and simplify:

\begin{align} \dfrac {\sqrt{3}} {2\sqrt{5}} &= \\[5pt] \dfrac {\sqrt{3}} {2\sqrt{5}} \cdot \dfrac {\sqrt{5}} {\sqrt{5}} &= \\[5pt] \dfrac {\sqrt{3}\cdot \sqrt{5}} {2\cdot ( \sqrt{5} )^2} &= \\[5pt] \dfrac {\sqrt{3}\cdot \sqrt{5}} {2\cdot 5}&= \\[5pt] \dfrac {\sqrt{3}\cdot \sqrt{5}} {10}& \end{align}

Now, we use the product rule to simplify the numerator:

\begin{align} \dfrac {\sqrt{3}\cdot \sqrt{5}} {10} &= \\[5pt] \dfrac {\sqrt{3\cdot5}} {10}&= \\[5pt] \dfrac {\sqrt{15}} {10}& \end{align}

To remove the radical $\sqrt{2}$ from the denominator, all we need to do is multiply both the numerator and denominator by $\sqrt{2}$ and simplify:

\begin{align} \dfrac {6\sqrt{3}} {\sqrt{2}} &= \\[5pt] \dfrac {6\sqrt{3}} {\sqrt{2}} \cdot \dfrac {\sqrt{2}} {\sqrt{2}} &= \\[5pt] \dfrac {6\sqrt{3}\cdot \sqrt{2}} {( \sqrt{2} )^2} &= \\[5pt] \dfrac {6\sqrt{3}\cdot \sqrt{2}} {2}&= \\[5pt] {3\sqrt{3}\cdot \sqrt{2}} & \end{align}

Now, we use the product rule:

\begin{align} {3\sqrt{3}\cdot \sqrt{2}} &= \\[5pt] 3{\sqrt{3\cdot 2}} &= \\[5pt] 3\sqrt{6}& \end{align}

To remove the radical $\sqrt{8}$ from the denominator, all we need to do is multiply both the numerator and denominator by $\sqrt{8}$ and simplify:

\begin{align} \dfrac {\sqrt{3}} {\sqrt{8}} &= \\[5pt] \dfrac {\sqrt{3}} {\sqrt{8}} \cdot \dfrac {\sqrt{8}} {\sqrt{8}} &= \\[5pt] \dfrac {\sqrt{3}\cdot\sqrt{8}} {( \sqrt{8} )^2} &= \\[5pt] \dfrac {\sqrt{3}\cdot\sqrt{8}} {8}& \end{align}

Next, we notice that $8$ has a perfect square factor $(4).$ Therefore, we can rewrite $\sqrt{8}$ as follows:

\begin{align*} \sqrt{8} &=\\[5pt] \sqrt{4\cdot 2} &=\\[5pt] \sqrt 4\cdot \sqrt 2 &=\\[5pt] 2\sqrt 2& \end{align*}

Finally, we simplify our original fraction as follows:

\begin{align} \dfrac {\sqrt{3}\cdot\sqrt{8}} {8} &= \\[5pt] \dfrac {\sqrt{3}\cdot 2\sqrt 2} {8} &= \\[5pt] \dfrac {2\cdot \sqrt{3}\cdot \sqrt 2} {8} &= \\[5pt] \dfrac {\sqrt{3}\cdot \sqrt 2} {4} &= \\[5pt] \dfrac {\sqrt{3\cdot 2}} {4} &= \\[5pt] \dfrac{\sqrt 6}{4} \end{align}

To remove the radical $\sqrt{3}$ from the denominator, all we need to do is multiply both the numerator and denominator by $\sqrt{3}$ and simplify:

\begin{align} \dfrac {\sqrt{8}} {\sqrt{3}} &= \\[5pt] \dfrac {\sqrt{8}} {\sqrt{3}} \cdot \dfrac {\sqrt{3}} {\sqrt{3}} &= \\[5pt] \dfrac {\sqrt{8}\cdot \sqrt{3}} {( \sqrt{3} )^2} &= \\[5pt] \dfrac {\sqrt{8}\cdot \sqrt{3}} {3}& \end{align}

Next, we notice that $8$ has a perfect square factor $(4).$ Therefore, we can rewrite $\sqrt{8}$ as follows:

\begin{align*} \sqrt{8} &=\\[5pt] \sqrt{4\cdot 2} &=\\[5pt] \sqrt 4\cdot \sqrt 2 &=\\[5pt] 2\sqrt 2& \end{align*}

Finally, we simplify our original fraction as follows:

\begin{align} \dfrac {\sqrt{8}\cdot \sqrt{3}} {3} &= \\[5pt] \dfrac {2\sqrt{2}\cdot \sqrt 3} {3} &= \\[5pt] \dfrac {2\sqrt{2\cdot 3}} {3} &= \\[5pt] \dfrac{2\sqrt 6}{3} \end{align}