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?

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
$$
\frac{\sqrt{7\sqrt{3}+4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}=\frac{(7\sqrt{3}+4\sqrt{5})\sqrt{67}}{67}
$$
