Odpowiedź:
[tex]\dfrac{2\sqrt{3}-1}{2-\sqrt{3}}=\dfrac{2\sqrt{3}-1}{2-\sqrt{3}}\cdot\dfrac{2+\sqrt{3}}{2+\sqrt{3}}=\dfrac{(2\sqrt{3}-1)(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}=\dfrac{4\sqrt{3}+2\cdot3-2-\sqrt{3}}{2^2-(\sqrt{3})^2}=\\\\\\=\dfrac{3\sqrt{3}+6-2-\sqrt{3}}{4-3}=\dfrac{3\sqrt{3}+4}{1}=3\sqrt{3}+4[/tex]
[tex]\dfrac{1}{\sqrt{7}-\sqrt{6}}=\dfrac{1}{\sqrt{7}-\sqrt{6}}\cdot\dfrac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}=\dfrac{\sqrt{7}+\sqrt{6}}{(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})}=\dfrac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^2-(\sqrt{6})^2}=\\\\\\=\dfrac{\sqrt{7}+\sqrt{6}}{7-6}=\dfrac{\sqrt{7}+\sqrt{6}}{1}=\sqrt{7}+\sqrt{6}[/tex]
[tex]\dfrac{3}{\sqrt{6}-\sqrt{3}}=\dfrac{3}{\sqrt{6}-\sqrt{3}}\cdot\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}}=\dfrac{3(\sqrt{6}+\sqrt{3})}{(\sqrt{6}-\sqrt{3})(\sqrt{6}+\sqrt{3})}=\dfrac{3(\sqrt{6}+\sqrt{3})}{(\sqrt{6}-\sqrt{3})(\sqrt{6}+\sqrt{3})}=\\\\\\=\dfrac{3(\sqrt{6}+\sqrt{3})}{(\sqrt{6})^2-(\sqrt{3})^2}=\dfrac{3(\sqrt{6}+\sqrt{3})}{6-3}=\dfrac{\not3(\sqrt{6}+\sqrt{3})}{\not3}=\sqrt{6}+\sqrt{3}[/tex]
[tex]\dfrac{4}{\sqrt{7}+\sqrt{3}}=\dfrac{4}{\sqrt{7} +\sqrt{3}}\cdot\dfrac{\sqrt{7}-\sqrt{3}}{\sqrt{7}-\sqrt{3} }=\dfrac{4(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}=\dfrac{4(\sqrt{7}-\sqrt{3})}{(\sqrt{7})^2-(\sqrt{3})^2}=\\\\\\=\dfrac{4(\sqrt{7}-\sqrt{3})}{7-3}=\dfrac{\not4(\sqrt{7}-\sqrt{3})}{\not4}=\sqrt{7}-\sqrt{3}[/tex]
