Fractions with radicals in the denominator I'm working my way through the videos on the Khan Academy, and have a hit a road block. I can't understand why the following is true:
$$\frac{6}{\quad\frac{6\sqrt{85}}{85}\quad} = \sqrt{85}$$

With respect to handling square roots in the denominator:
Say, for example, you have $\displaystyle \frac {a}{\sqrt{b}}$. 
If you want a "square-root free" denominator, remember that you can multiply any fraction by the number $1$ (or its equivalent) without changing its value.  In this case, we want to multiply the given fraction by $\displaystyle \frac{\sqrt{b}}{\sqrt{b}} = 1$. Can you see why we would want to choose this particular representation of 1?  
Now, $$\frac{\sqrt{b}}{\sqrt{b}}\cdot \frac {a}{\sqrt{b}} = \frac{a\cdot \sqrt{b}}{(\sqrt{b})^2} = \frac{a\cdot \sqrt{b}}{b}$$
And if the greatest common divisor of a and b > 1, then we can reduce $\displaystyle \frac ab$ by dividing each of $a$ and $b$ by their greatest common divisor (i.e., reducing the fraction $\displaystyle \frac ab$.
Now, we take a look at a slightly more complicated situation, as in your problem, where the denominator is itself a fraction given by $\displaystyle \frac{6}{\frac{6\sqrt{85}}{85}}$
We can first multiply numerator and denominator by 85, then proceed as we did in the case above: $$\frac{6}{\frac{6\sqrt{85}}{85}}\cdot \frac{85}{85} = \frac{6\cdot 85}{6\cdot\sqrt{85}} = \frac {85}{\sqrt{85}} = \frac {85}{\sqrt{85}}\cdot \frac{\sqrt{85}}{\sqrt{85}} = \frac{85\cdot \sqrt{85}}{85} = \sqrt{85} $$
More simply, once we have reduced the expression to $\displaystyle \frac{85}{\sqrt{85}}$, we notice that $85 = (\sqrt{85})^2$, and so, this, together with the fact that we can then cancel a common factor of both the numerator and the denominator, we have:
$$\frac{85}{\sqrt{85}} = \frac{(\sqrt{85})^2}{\sqrt{85}} = \sqrt{85}$$ 
Note: When you have enough experience simplifying fractions and fractions with fractions with roots, etc., many of the steps which I've displayed explicitly above will be "second nature" for you, and you'll be able to manipulate such expressions with far less in the way of computations than I've shown above.
