Question

$Alguem? \frac{ \frac{3}{4}+ \sqrt{3} }{1+ \sqrt{3} }$

Answer 1

$\frac{\frac{3}{4}+\sqrt3}{1+\sqrt3}.\frac{1-\sqrt3}{1-\sqrt3}=\frac{\frac{3}{4}-\frac{3\sqrt3}{4}+\sqrt3-3}{1-3}=\\ \\ \frac{-\frac{9}{4}+\frac{\sqrt3}{4}}{-2}=\frac{\frac{\sqrt3-9}{4}}{-2}=-\frac{\sqrt3-9}{8}$