Odpowiedź:
a)
[tex]\frac{2\sqrt{6}-1}{\sqrt{6}+3} - \sqrt{32\frac{2}{3}} = \frac{2\sqrt{6}-1}{\sqrt{6}+3} -\sqrt{\frac{98}{3}} =\frac{2\sqrt{6}-1}{\sqrt{6}+3}-\frac{7\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{3}(2\sqrt{6}-1)-7\sqrt{2}(\sqrt{6} +3)}{(\sqrt{6}+3)*\sqrt{3}} =\frac{2\sqrt{18}-\sqrt{3}-7\sqrt{12} -21\sqrt{2}}{\sqrt{18}+3\sqrt{3}}=\frac{6\sqrt{2}-\sqrt{3}-14\sqrt{3}-21\sqrt{2}}{3\sqrt{2}+3\sqrt{3}} =\frac{-15\sqrt{3}-15\sqrt{2}}{3\sqrt{2}+3\sqrt{3}}=\frac{-5(3\sqrt{3}+3\sqrt{2})}{3\sqrt{3}+3\sqrt{2}} =-5[/tex]
b)
[tex]\frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}+\sqrt{3}} -\frac{\sqrt{3}+\sqrt{6}}{\sqrt{3}-\sqrt{6}} =\frac{(\sqrt{6}-\sqrt{3})(\sqrt{3}-\sqrt{6})-(\sqrt{3}+\sqrt{6})^{2}}{(\sqrt{6}+\sqrt{3})(\sqrt{3}-\sqrt{6})}=\frac{\sqrt{18}-6-3+\sqrt{18}-3-2\sqrt{18}-6}{3-6} =\frac{-18}{-3} =6[/tex]
Szczegółowe wyjaśnienie:
