Odpowiedź:
[tex]\dfrac{3\sqrt{2}}{2\sqrt{2}+1}=\dfrac{3\sqrt{2}}{2\sqrt{2}+1}\cdot\dfrac{2\sqrt{2}-1}{2\sqrt{2}-1}=\dfrac{3\sqrt{2}(2\sqrt{2}-1)}{(2\sqrt{2}+1)(2\sqrt{2}-1)}=\dfrac{6\cdot2-3\sqrt{2}}{(2\sqrt{2})^2-1^2}=\\\\\\=\dfrac{12-3\sqrt{2}}{4\cdot2-1}=\dfrac{12-3\sqrt{2}}{8-1}=\dfrac{12-3\sqrt{2}}{7}\\\\\\Odp.B[/tex]
