As promised, here lie the \emph{binomial denominators}. There are a couple of prerequisites to understanding this mathematics, the first is that you must totally grasp $(a+b)(a-b) = a^2 - b^2$.

The significance of this result is that if we make $a$ and $b$ surds like so: $(\sqrt{a}+\sqrt{b})$, then multiplying by $(\sqrt{a}-\sqrt{b})$ yields a fantastic result, namely:

\vspace{0.5cm}
\begin{equation}
    (\source{\sqrt{a}}+\source{\sqrt{b}})(\target{\sqrt{a}}-\target{\sqrt{b}})= (\sqrt{a})^2 - (\sqrt{b})^2\mbox{\drawarrows}\\
    = a - b 
\end{equation}
\vspace{0.5cm}

This act of starting with $\sqrt{a} + \sqrt{b}$, and multiplying it with the sign flipped: $\sqrt{a}-\sqrt{b}$ is called \emph{multiplying by the reciprocal}. Let's practise this method:

 \begin{examplebox}                                                                                                                   
     \subsection{Examples:}                                                                                                           
     \begin{questions}                                                                                                                
         \Question[1] \[\frac{2 \sqrt{5}}{2 \sqrt{5}-2}=\]                                                                            
             \begin{solutionordottedlines}[3cm]                                                                                       
                 Multiply by the conjugate: $\frac{2 \sqrt{5}}{2 \sqrt{5}-2} \cdot \frac{2 \sqrt{5}+2}{2 \sqrt{5}+2} = \frac{4 \cdot  
 + 4 \sqrt{5}}{20 - 4} = \frac{20 + 4 \sqrt{5}}{16} = \frac{5}{4} + \sqrt{5}$                                                         
             \end{solutionordottedlines}                                                                                              
         \Question[1] \[\frac{\sqrt{3}+\sqrt{2}}{3 \sqrt{2}+2 \sqrt{3}}\]                                                             
             \begin{solutionordottedlines}[3cm]                                                                                       
                 Multiply by the conjugate: $\frac{\sqrt{3}+\sqrt{2}}{3 \sqrt{2}+2 \sqrt{3}} \cdot \frac{3 \sqrt{2}-2 \sqrt{3}}{3     
 \sqrt{2}-2 \sqrt{3}} = \frac{3 \cdot 2 - 2 \cdot 3}{6 - 4 \sqrt{6} + 4 \sqrt{6} - 12} = \frac{6 - 6}{-6} = 0$                        
             \end{solutionordottedlines}                                                                                              
     \end{questions}                                                                                                                  
 \end{examplebox}                                                                                                                     


 \begin{exercisebox}                                                                                                                  
     \subsection{Exercises:}                                                                                                          
     \begin{questions}                                                                                                                
         \Question[6]                                                                                                                 
         \begin{parts}\begin{multicols}{2}                                                                                            
             \part \(\frac{3}{\sqrt{2}-1}\)                                                                                           
                 \begin{solutionordottedlines}[3cm]                                                                                   
                     Multiply by the conjugate: $\frac{3}{\sqrt{2}-1} \cdot \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{3(\sqrt{2}+1)}{2-1} 
 3\sqrt{2} + 3$                                                                                                                       
                 \end{solutionordottedlines}                                                                                          
             \part \(\frac{2}{3+\sqrt{5}}\)                                                                                           
                 \begin{solutionordottedlines}[3cm]                                                                                   
                     Multiply by the conjugate: $\frac{2}{3+\sqrt{5}} \cdot \frac{3-\sqrt{5}}{3-\sqrt{5}} = \frac{6 - 2\sqrt{5}}{9} - = \frac{6 - 2\sqrt{5}}{4} = \frac{3}{2} - \frac{\sqrt{5}}{2}$                                                                        
                 \end{solutionordottedlines}                                                                                          
             \part \(\frac{4}{\sqrt{5}+\sqrt{2}}\)                                                                                    
                 \begin{solutionordottedlines}[3cm]                                                                                   
                     Multiply by the conjugate: $\frac{4}{\sqrt{5}+\sqrt{2}} \cdot \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} =      
 \frac{4\sqrt{5} - 4\sqrt{2}}{5 - 2} = \frac{4\sqrt{5} - 4\sqrt{2}}{3}$                                                               
                 \end{solutionordottedlines}                                                                                          
             \part \(\frac{1}{\sqrt{7}-\sqrt{5}}\)                                                                                    
                 \begin{solutionordottedlines}[3cm]                                                                                   
                     Multiply by the conjugate: $\frac{1}{\sqrt{7}-\sqrt{5}} \cdot \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}} =      
 \frac{\sqrt{7}+\sqrt{5}}{7 - 5} = \frac{\sqrt{7}+\sqrt{5}}{2}$                                                                       
                 \end{solutionordottedlines}                                                                                          
             \part \(\frac{1}{\sqrt{2}-1}\)                                                                                           
                 \begin{solutionordottedlines}[3cm]                                                                                   
                     Multiply by the conjugate: $\frac{1}{\sqrt{2}-1} \cdot \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{\sqrt{2}+1}{2-1} =  
 \sqrt{2} + 1$                                                                                                                        
                 \end{solutionordottedlines}                                                                                          
             \part \(\frac{1}{\sqrt{3}-\sqrt{2}}\)                                                                                    
                 \begin{solutionordottedlines}[3cm]                                                                                   
                     Multiply by the conjugate: $\frac{1}{\sqrt{3}-\sqrt{2}} \cdot \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} =      
 \frac{\sqrt{3}+\sqrt{2}}{3 - 2} = \sqrt{3} + \sqrt{2}$                                                                               
                 \end{solutionordottedlines}                                                                                          
         \end{multicols}\end{parts}                                                                                                   
     \end{questions}                                                                                                                  
 \end{exercisebox}