how to simplify $\frac{2+\sqrt{3}}{6+4\sqrt{3}}$ among solving a problem I need to simplify: $$\frac{2+\sqrt{3}}{6+4\sqrt{3}}$$
I know it is $\frac{1}{2\sqrt{3}}$ but I dont know how to show that in mathematic way for example I can write $$\frac{2+\sqrt{3}}{2\sqrt{3}(\sqrt{3}+2)}$$ but still not obvious.Is there better way to prove/show that?
 A: If in doubt, use conjugates:
$$ \frac{2+\sqrt 3}{6+4\sqrt 3}=\frac{2+\sqrt 3}{2\cdot (3+2\sqrt 3)}=\frac{(2+\sqrt 3)(3-2\sqrt 3)}{2\cdot (3+2\sqrt 3)(3-2\sqrt 3)}=\frac{-\sqrt 3}{-6}=\frac1{2\sqrt 3}$$
A: The general approach for something like this would be to multiply by the conjugate of the denominator. More explicitly, say you have
$$\frac{a + b \sqrt c}{x + y \sqrt c}$$
Then you multiply the top and bottom by $x - y \sqrt c $ (which we call the "conjugate" of $x+y \sqrt c$ in this sort of context):
$$\frac{a + b \sqrt c}{x + y \sqrt c} \cdot \frac{x-y \sqrt c}{x-y \sqrt c}$$ 
Simplifying leads to
$$\frac{(ax - byc) + (bx-ay) \sqrt c}{x^2 - cy^2}$$
The motivating reason for this approach is the difference of squares identity $(p-q)(p+q) = p^2 - q^2$, which makes simplifying the denominator easier.
The conjugate to multiply by your expression's top and bottomin your case is $6-4 \sqrt 3$. (However your factoring method is equally valid, arguably even easier!)
A: We have that
$$\frac{2+\sqrt{3}}{6+4\sqrt{3}}=\frac{2+\sqrt{3}}{6+4\sqrt{3}}\cdot \frac{6-4\sqrt{3}}{6-4\sqrt{3}}=\frac{-2\sqrt{3}}{-12}=\frac{\sqrt 3}6$$
or by your idea
$$\frac{2+\sqrt{3}}{6+4\sqrt{3}}=\frac{2+\sqrt{3}}{2\sqrt{3}(\sqrt{3}+2)}=\frac1{2\sqrt{3}}=\frac1{2\sqrt{3}}\cdot\frac{2\sqrt{3}}{2\sqrt{3}}=\frac{\sqrt 3}6$$
