Odpowiedź:
[tex]\dfrac{3}{2\sqrt{3}-1}=\dfrac{3}{2\sqrt{3}-1}\cdot\dfrac{2\sqrt{3}+1}{2\sqrt{3}+1}=\dfrac{3(2\sqrt{3}+1)}{(2\sqrt{3}-1)(2\sqrt{3}+1)}=\dfrac{6\sqrt{3}+3}{(2\sqrt{3})^2-1^2}=\\\\\\=\dfrac{6\sqrt{3}+3}{4\cdot3-1}=\dfrac{6\sqrt{3}+3}{12-1}=\dfrac{6\sqrt{3}+3}{11}[/tex]
