[tex]\displaystyle\\
\frac{1}{2+\sqrt{3}}-\frac{1}{2-\sqrt{3}}=~~\text{(Rationalizam numitorii.)}\\\\
=\frac{1\times(2-\sqrt{3})}{(2+\sqrt{3})\times(2-\sqrt{3})}-\frac{1\times(2+ \sqrt{3})}{(2-\sqrt{3})\times (2+\sqrt{3})}=\\\\
=\frac{2-\sqrt{3}}{2^2-\Big(\sqrt{3}\Big)^2 }-\frac{2+\sqrt{3}}{2^2-\Big(\sqrt{3}\Big)^2 }=\\\\
=\frac{2-\sqrt{3}}{ 4-3}-\frac{2+\sqrt{3}}{4-3}=\\\\
=\frac{2-\sqrt{3}}{1}-\frac{2+\sqrt{3}}{1}=\\\\
=2-\sqrt{3}-(2+\sqrt{3})=2-\sqrt{3}-2-\sqrt{3}=\boxed{\bf -2\sqrt{3}}[/tex]
Daca intre fractii ar fi fost semnul "+" atunci s-ar fi redus radicalii si rezultatul final ar fi fost 4 ∈ N.
