Odpowiedź:
[tex]a)\ \ \dfrac{8}{3\sqrt{2}-4}=\dfrac{8}{3\sqrt{2}-4}\cdot\dfrac{3\sqrt{2}+4}{3\sqrt{2}+4}=\dfrac{8(3\sqrt{2}+4)}{(3\sqrt{2}-4)(3\sqrt{2}+4)}=\dfrac{8(3\sqrt{2}+4)}{(3\sqrt{2})^2-4^2}=\\\\\\=\dfrac{8(3\sqrt{2}+4)}{9\cdot2-16}=\dfrac{8(3\sqrt{2}+4)}{18-16}=\dfrac{\not8^4(3\sqrt{2}+4)}{\not2_{1}}=4(3\sqrt{2}+4)=12\sqrt{2}+16[/tex]
[tex]b)\ \ \dfrac{6}{3+2\sqrt{3}}=\dfrac{6}{3+2\sqrt{3}}\cdot\dfrac{3-2\sqrt{3}}{3-2\sqrt{3}}=\dfrac{6(3-2\sqrt{3})}{(3+2\sqrt{3})(3-2\sqrt{3})}=\dfrac{6(3-2\sqrt{3})}{3^2-(2\sqrt{3})^2}=\\\\\\=\dfrac{6(3-2\sqrt{3})}{9-4\cdot3}=\dfrac{6(3-2\sqrt{3})}{9-12}=\dfrac{\not6^2(3-2\sqrt{3})}{-\not3_{1}}=-2(3-2\sqrt{3})=-6+4\sqrt{3}[/tex]
[tex]c)\ \ \dfrac{2\sqrt{3}-1}{2\sqrt{3}+2}=\dfrac{2\sqrt{3}-1}{2\sqrt{3}+2}\cdot\dfrac{2\sqrt{3}-2}{2\sqrt{3}-2}=\dfrac{(2\sqrt{3}-1)(2\sqrt{3}-2)}{(2\sqrt{3}+2)(2\sqrt{3}-2)}=\dfrac{4\cdot3-4\sqrt{3}-2\sqrt{3}+2}{(2\sqrt{3})^2-2^2}=\\\\\\=\dfrac{12-6\sqrt{3}+2}{4\cdot3-4}=\dfrac{14-6\sqrt{3}}{12-4}=\dfrac{14-6\sqrt{3}}{8}=\dfrac{\not2^1(7-3\sqrt{3})}{\not8_{4}}=\dfrac{7-3\sqrt{3}}{4}[/tex]
a)
[tex]\dfrac{8}{3\sqrt{2}-4} =\dfrac{8}{3\sqrt{2}-4}\cdot\dfrac{3\sqrt{2}+4}{3\sqrt{2}+4} = \dfrac{8(3\sqrt{2}+4)}{(3\sqrt{2}-4)(3\sqrt{2}+4)} = \dfrac{8(3\sqrt{2}+4)}{18-16}=\dfrac{8(3\sqrt{2}+4)}{2} =\\\\=4(3\sqrt{2}+4)=12\sqrt{2}+16[/tex]
b)
[tex]\dfrac{6}{3+2\sqrt{3}}=\dfrac{6}{3+2\sqrt{3}}\cdot\dfrac{3-2\sqrt{3}}{3-2\sqrt{3}} = \dfrac{6(3-2\sqrt{3})}{(3+2\sqrt{3})(3-2\sqrt{3})} =\dfrac{6(3-2\sqrt{3})}{9-12} = \dfrac{6(3-2\sqrt{3})}{-3} =\\\\= 2(2\sqrt{3}-3)=4\sqrt{3}-6[/tex]
c)
[tex]\dfrac{2\sqrt{3}-1}{2\sqrt{3}+2} =\dfrac{2\sqrt{3}-1}{2\sqrt{3}+2}\dcdot\frac{2\sqrt{3}-2}{2\sqrt{3}-2} = \dfrac{(2\sqrt{3}-1)(2\sqrt{3}-2)}{(2\sqrt{3}+2)(2\sqrt{3}-2)} = \dfrac{12-4\sqrt{3}-2\sqrt{3}+2}{12-4}=\\\\\\=\dfrac{12-4\sqrt{3}-2\sqrt{3}+2}{8}=\dfrac{14-6\sqrt{3}}{8} = \dfrac{2(7-3\sqrt{3})}{8}=\dfrac{7-3\sqrt{3}}{4}[/tex]
