[tex]\bf \cfrac{8-\sqrt{18}}{\sqrt{2}}=a+b\sqrt{2}\qquad \qquad 18\to 3\cdot 3\cdot 2\to 3^2\cdot 2\\\\
-----------------------------\\\\
\cfrac{8-\sqrt{3^2\cdot 2}}{\sqrt{2}}\implies \cfrac{8-3\sqrt{2}}{\sqrt{2}}\impliedby \textit{now, let's distribute the denominator}
\\\\\\
\cfrac{8}{\sqrt{2}}-\cfrac{3\sqrt{2}}{\sqrt{2}}\implies \boxed{\cfrac{8}{\sqrt{2}}-3}\impliedby \textit{let's rationalize some}\\\\
-----------------------------\\\\[/tex]
[tex]\bf \cfrac{8}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{8\sqrt{2}}{(\sqrt{2})^2}\implies \cfrac{8\sqrt{2}}{2}\implies 4\sqrt{2}\\\\
-----------------------------\\\\
thus\qquad \cfrac{8}{\sqrt{2}}-3\iff 4\sqrt{2}-3\implies
\begin{array}{ccllll}
-3&+4&\sqrt{2}\\
\uparrow &\uparrow \\
a&b
\end{array}[/tex]
