Write this surd in its simplest form. 
Express $\dfrac{1}{2+ \sqrt3}$ in its simplest form.

NB: The textbook has the answer as $2 - \sqrt3$ but I can't see how that was achieved.
I tried $\dfrac{1}{2} + \dfrac{1}{\sqrt3}$ and multiplying the top and bottom by $\sqrt3 $ to get $\dfrac{1}{2} + \dfrac{\sqrt3}{3}$ so far.

Multiply by $\displaystyle 1 = \frac{2-\sqrt 3}{2-\sqrt 3}$. Use $(a-b)(a+b) = a^2-b^2$ on the denominator. It's called "rationalising the denominator" by multiplying the denominator by the "conjugate surd". You should look up the key phrases in quotes.
