Rationalize the denominator of $\frac{\sqrt{7\sqrt{3}+4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}$ I have to rationalize the denominator of $A = 
\frac{\sqrt{7\sqrt{3}+4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}$. I multiplied the fraction by $\frac{\sqrt{7\sqrt{3}-4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}$, and $A=\frac{\sqrt{67}}{7\sqrt{3}-4\sqrt{5}}$. Then I multiplied by $\frac{7\sqrt{3}+4\sqrt{5}}{7\sqrt{3}+4\sqrt{5}}$, and $A=\frac{7\sqrt{201}+4\sqrt{335}}{67}$. Can someone tell me if there is a better solution and whether I am right?
 A: Hint: Multiply numerator and denominator by $$\sqrt{7\sqrt{3}+4\sqrt{5}}$$ and then by $$\sqrt{67}$$
For your work we get
$$\frac{(7\sqrt{3}+4\sqrt{5})\sqrt{67}}{67}$$
A: 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
$\dfrac{1}{\sqrt{2\sqrt{21}-4\sqrt{5}}}=\dfrac{\sqrt{2\sqrt{21}-4\sqrt{5}}}{2\sqrt{21}-4\sqrt{5}}$
A: For positive $a$ and $b$, we have $\sqrt{a}/\sqrt{b}=\sqrt{a/b}$. So you can concentrate on
$$
\frac{7\sqrt{3}+4\sqrt{5}}{7\sqrt{3}-4\sqrt{5}}=\frac{(7\sqrt{3}+4\sqrt{5})^2}{49\cdot3-15\cdot5}=\frac{(7\sqrt{3}+4\sqrt{5})^2}{67}
$$
Thus you're done with
$$
\frac{\sqrt{7\sqrt{3}+4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}=\frac{(7\sqrt{3}+4\sqrt{5})\sqrt{67}}{67}
$$
