<s>[INST] 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?
 [/INST] 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
$$\frac{2+\sqrt{3}}{6+4\sqrt{3}}=\frac{2+\sqrt{3}}{2\sqrt{3}(\sqrt{3}+2)}=\frac1{2\sqrt{3}}=\frac1{2\sqrt{3}}\cdot\frac{2\sqrt{3}}{2\sqrt{3}}=\frac{\sqrt 3}6$$
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