+2666 We have: \(\large{1 \over {1 + {1 \over {\sqrt3 + 2}}}}\)
Lets start with \(\large{1 \over {\sqrt3 + 2}}\). To rationalize the denominator, multiply the fraction by \(\large{{ \sqrt 3 - 2} \over { \sqrt 3 - 2}}\). By doing this, we keep the value of the fraction the same, while keeping square roots out of the denominator, because we are, effectivelty, multiplying by 1. The denominator of the fraction becomes \({(\sqrt 3 + 2) ( \sqrt 3 - 2) = 3 - 4 = -1}\). However, the numerator is still \(\large{\sqrt 3 -2} \), because anything multiplied by 1 is itself.
Now, the fraction is: \(\large {{ \sqrt 3 - 2 } \over -1}\). Multiplying by \(\large{ -1 \over -1}\), we get: \(2 - \sqrt 3 \) remember, we aren't changing the value of the fraction by doing this, because \({ -1 \over -1} =1\).
Adding 1 to this gives us \(3 - \sqrt 3 \).
This means we have the fraction \(\large{1\over { 3 - \sqrt 3}}\).
Now, we have to rationalize this again.
Can you take it from here?
