a)[tex]2(3\sqrt{3}-\sqrt{2})-3(3\sqrt{3}+\sqrt{2})=6\sqrt{3}-2\sqrt{2}-9\sqrt{3}-3\sqrt{2}=[/tex]
[tex]=-3\sqrt{3}-5\sqrt{2}[/tex]
b)[tex]\frac{1}{3\sqrt{3}-\sqrt{2}}+\frac{1}{3\sqrt{3}+\sqrt{2}}=[/tex]
[tex]=\frac{3\sqrt{3}+\sqrt{2}+3\sqrt{3}-\sqrt{2}}{25}=\frac{6\sqrt{3}}{25}[/tex]
Asta a fost suma inverselor
[tex]3\sqrt{3}-\sqrt{2}+3\sqrt{3}+\sqrt{2}=6\sqrt{3}[/tex]
Inversul sumei este:
[tex]\frac{1}{6\sqrt{3}}=\frac{\sqrt{3}}{18}[/tex]
