Rationalize the denominator of $\frac{\sqrt{7\sqrt{3}+4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}$
Hint: Multiply numerator and denominator by $$\sqrt{7\sqrt{3}+4\sqrt{5}}$$ and then by $$\sqrt{67}$$ For your work we get $$\frac{(7\sqrt{3}+4\sqrt{5})\sqrt{67}}{67}$$
Various others have suggested mulrltiplying by $\sqrt{7\sqrt{3}\color{blue}{+}4\sqrt{5}}$ in the first step. The reason this is preferable is because you get a product with at most one square root in the denominator, which simplifies the computation; thus
$\dfrac{7\sqrt{3}+4\sqrt{5}}{\color{blue}{\sqrt{67}}}$
versus your expression
$\dfrac{\sqrt{67}}{7\sqrt{3}-4\sqrt{5}}$
The suggested alternative has a hidden advantage. Though not in this case, sometimes the remaining square root has a squared quantity in the denominator allowing a step to be saved. For instance, compare
$\dfrac{1}{\sqrt{2\sqrt{21}-4\sqrt{5}}}=\dfrac{\sqrt{2\sqrt{21}+4\sqrt{5}}}{\sqrt{4}}=\dfrac{\sqrt{2\sqrt{21}+4\sqrt{5}}}{2}$
with
$\dfrac{1}{\sqrt{2\sqrt{21}-4\sqrt{5}}}=\dfrac{\sqrt{2\sqrt{21}-4\sqrt{5}}}{2\sqrt{21}-4\sqrt{5}}$