[tex]\frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\\\\\\\frac{3+\sqrt{7}}{5\sqrt{7}} = \frac{3+\sqrt{7}}{5\sqrt{7}}\cdot\frac{\sqrt{7}}{\sqrt{7}} =\frac{\sqrt{7}(3+\sqrt{7})}{(5\sqrt{7}\cdot\sqrt{7})} = \frac{3\sqrt{7}+7}{35}[/tex]
[tex]\frac{6+\sqrt{2}}{2+\sqrt{2}} = \frac{6+\sqrt{2}}{2+\sqrt{2}}\cdot\frac{2-\sqrt{2}}{2-\sqrt{2}}=\frac{(6+\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}=\frac{12-6\sqrt{2}+2\sqrt{2}-2}{4-2} = \frac{10-4\sqrt{2}}{2} = 5-2\sqrt{2}[/tex]
[tex]\frac{3\sqrt{2}+3}{3-2\sqrt{3}} = \frac{3\sqrt{2}+3}{3-2\sqrt{3}}\cdot\frac{3+2\sqrt{3}}{3+2\sqrt{3}} = \frac{(3\sqrt{2}+3)(3+2\sqrt{3})}{(3-2\sqrt{3})(3+2\sqrt{3})} =\frac{9\sqrt{2}+6\sqrt{6}+9+6\sqrt{3}}{9-12}=\frac{9\sqrt{2}+6\sqrt{6}+9+6\sqrt{3}}{-3}=\\\\=\frac{-3(-3\sqrt{2}-2\sqrt{6}-3-2\sqrt{3})}{-3} =-3\sqrt{2}-2\sqrt{6}}-3-2\sqrt{3}[/tex]
