Odpowiedź:
[tex]a)\ \ \dfrac{\sqrt{5}+1}{\sqrt{5}-1}=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}*\dfrac{\sqrt{5}+1}{\sqrt{5}+1}=\dfrac{(\sqrt{5}+1)(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\dfrac{(\sqrt{5}+1)^2}{(\sqrt{5})^2-1^2}=\\\\\\=\dfrac{5+2\sqrt{5}+1}{5-1}=\dfrac{6+2\sqrt{5}}{4}=\dfrac{\not2^1(3+\sqrt{5})}{\not4_{2}}=\dfrac{3+\sqrt{5}}{2}[/tex]
[tex]b)\ \ \dfrac{7\sqrt{2}}{2\sqrt{2}-3}=\dfrac{7\sqrt{2}}{2\sqrt{2}-3}\cdot\dfrac{2\sqrt{2}+3}{2\sqrt{2}+3}=\dfrac{7\sqrt{2}(2\sqrt{2}+3)}{(2\sqrt{2}-3)(2\sqrt{2}+3)}=\dfrac{14\cdot2+21\sqrt{2}}{(2\sqrt{2})^2-3^2}=\\\\\\=\dfrac{28+21\sqrt{2}}{4\cdot2-9}=\dfrac{28+21\sqrt{2}}{8-9}=\dfrac{28+21\sqrt{2}}{-1}=-(28+21\sqrt{2})=-28-21\sqrt{2}[/tex]
[tex]c)\ \ \dfrac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}=\dfrac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\cdot\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}=\dfrac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}=\dfrac{\sqrt{12}-\sqrt{18}}{(\sqrt{2})^2-(\sqrt{3})^2}=\\\\\\=\dfrac{\sqrt{4\cdot3}-\sqrt{9\cdot2}}{2-3}=\dfrac{2\sqrt{3}-3\sqrt{2}}{-1}=-(2\sqrt{3}-3\sqrt{2})=-2\sqrt{3}+3\sqrt{2}[/tex]
[tex]d)\ \ \dfrac{\sqrt{3}-1}{2\sqrt{3}+\sqrt{2}}=\dfrac{\sqrt{3}-1}{2\sqrt{3}+\sqrt{2}}\cdot\dfrac{2\sqrt{3}-\sqrt{2}}{2\sqrt{3}-\sqrt{2}}=\dfrac{(\sqrt{3}-1)(2\sqrt{3}-\sqrt{2}) }{(2\sqrt{3}+\sqrt{2})(2\sqrt{3}-\sqrt{2})}=\\\\\\=\dfrac{2\cdot3-\sqrt{6}-2\sqrt{3}+\sqrt{2}}{(2\sqrt{3})^2-(\sqrt{2})^2}=\dfrac{6-\sqrt{6}-2\sqrt{3}+\sqrt{2} }{4\cdot3-2}=\dfrac{6-\sqrt{6}-2\sqrt{3}+\sqrt{2}}{12-2}=\\\\\\=\dfrac{6-\sqrt{6}-2\sqrt{3}+\sqrt{2}}{10}[/tex]
