Odpowiedź:
[tex]a=3\sqrt{6}-5\ \ \ \ i\ \ \ \ b=\sqrt{12}-2\sqrt{2}\\\\\\a^2-\sqrt{3}b=(3\sqrt{6}-5)^2-\sqrt{3}(\sqrt{12}-2\sqrt{2})=(3\sqrt{6})^2-2\cdot3\sqrt{6}\cdot5+5^2-\sqrt{36}+2\sqrt{6}=\\\\=9\cdot6-30\sqrt{6}+25-6+2\sqrt{6}=54-28\sqrt{6}+19=73-28\sqrt{6}[/tex]
