Odpowiedź:
[tex]a=2-\sqrt{3}\ \ \ \ i\ \ \ \ b=3+2\sqrt{3}\\\\\\a-b=2-\sqrt{3}-(3+2\sqrt{3})=2-\sqrt{3}-3-2\sqrt{3}=-1-3\sqrt{3}\\\\\\a\cdot b=(2-\sqrt{3})(3+2\sqrt{3})=6+4\sqrt{3}-3\sqrt{3}-2\cdot3=6+\sqrt{3}-6=\sqrt{3}\\\\\\\dfrac{a}{b}=\dfrac{2-\sqrt{3}}{3+2\sqrt{3}}=\dfrac{2-\sqrt{3}}{3+2\sqrt{3}}\cdot\dfrac{3-2\sqrt{3}}{3-2\sqrt{3}}=\dfrac{(2-\sqrt{3})(3-2\sqrt{3})}{(3+2\sqrt{3})(3-2\sqrt{3})}=\dfrac{6-4\sqrt{3}-3\sqrt{3}+2\cdot3}{3^2-(2\sqrt{3})^2}=[/tex]
[tex]=\dfrac{6-7\sqrt{3}+6}{9-4\cdot3}=\dfrac{12-7\sqrt{3}}{9-12}=\dfrac{12-7\sqrt{3}}{-3}=-\dfrac{12-7\sqrt{3}}{3}\\\\\\\\\dfrac{1}{a}=\dfrac{1}{2-\sqrt{3}}=\dfrac{1}{2-\sqrt{3}}\cdot\dfrac{2+\sqrt{3}}{2+\sqrt{3}}=\dfrac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}=\dfrac{2+\sqrt{3}}{2^2-(\sqrt{3})^2}=\\\\\\=\dfrac{2+\sqrt{3}}{4-3}=\dfrac{2+\sqrt{3}}{1}=2+\sqrt{3}[/tex]
[tex]b^2=(3+2\sqrt{3})^2=3^2+2\cdot3\cdot2\sqrt{3}+(2\sqrt{3})^2=9+12\sqrt{3}+4\cdot3=9+12\sqrt{3}+12=\\\\=21+12\sqrt{3}[/tex]
