a) 2[tex]\sqrt{3\\}[/tex] ([tex]\sqrt{27} +3\sqrt{6} - 2\sqrt{3}[/tex])
b) [tex]\sqrt{3} (\sqrt{6} - \sqrt{2} ) + \frac{1}{\sqrt{6} } (\sqrt{3} - 2\sqrt{2} )[/tex]
PROSZE SZYBKOOO 40 PUNKTÓWWWWWWW

[tex]2\sqrt{3} (\sqrt{27} +3\sqrt{6}-2\sqrt{3}) = 2\sqrt{3} (3\sqrt{3} +3\sqrt{6}-2\sqrt{3}) = 2\sqrt{3} (\sqrt{3} +3\sqrt{2*3}) = 2\sqrt{3*3} +6\sqrt{3*3*2} = 6 + 18\sqrt{2}[/tex]

b)

[tex]\sqrt{3} (\sqrt{6} - \sqrt{2})+\frac{1}{\sqrt{6}} (\sqrt{3} - 2\sqrt{2}) = \sqrt{3} (\sqrt{2*3} - \sqrt{2})+\frac{1}{\sqrt{2*3}} (\sqrt{3} - 2\sqrt{2}) = \sqrt{2*3*3} - \sqrt{6}+\frac{\sqrt{3} }{\sqrt{2*3}} - \frac{2\sqrt{2} }{\sqrt{2*3}} = 3\sqrt{2} - \sqrt{6}+\frac{1 }{\sqrt{2}} - \frac{2}{\sqrt{3}} = 3\sqrt{2} - \sqrt{6}+\frac{\sqrt{2}}{2} - \frac{2\sqrt{3}}{3} = \frac{7\sqrt{2}}{2}- \frac{2\sqrt{3}}{3} - \sqrt{6}[/tex]

Magdaviz Magdaviz · Answer 2

Odpowiedź:

[tex]a)\ \ 2\sqrt{3}(\sqrt{27}+3\sqrt{6}-2\sqrt{3})=2\sqrt{3}\cdot\sqrt{27}+2\sqrt{3}\cdot3\sqrt{6}-2\sqrt{3}\cdot2\sqrt{3}=\\\\=2\sqrt{81}+6\sqrt{18}-4\sqrt{9}=2\cdot9+6\sqrt{9\cdot2}-4\cdot3=18+6\cdot3\sqrt{2}-12=6+18\sqrt{2}[/tex]

[tex]b)\ \ \sqrt{3}(\sqrt{6}-\sqrt{2})+\frac{1}{\sqrt{6}}(\sqrt{3}-2\sqrt{2})=\sqrt{18}-\sqrt{6}+\frac{\sqrt{3}}{\sqrt{6}}-\frac{2\sqrt{2}}{\sqrt{6}}=\\\\=\sqrt{9\cdot2}-\sqrt{6}+\frac{\sqrt{3}-2\sqrt{2}}{\sqrt{6}}=3\sqrt{2}-\sqrt{6}+\frac{\sqrt{3}-2\sqrt{2}}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}}=3\sqrt{2}-\sqrt{6}+\frac{(\sqrt{3}-2\sqrt{2})\cdot\sqrt{6}}{6}=[/tex]

[tex]=3\sqrt{2}-\sqrt{6}+\frac{\sqrt{18}-2\sqrt{12}}{6}=3\sqrt{2}-\sqrt{6}+\frac{\sqrt{9\cdot2}-2\sqrt{4\cdot3}}{6}=3\sqrt{2}-\sqrt{6}+\frac{3\sqrt{2}-2\cdot2\sqrt{3} }{6}=\\\\=3\sqrt{2}-\sqrt{6}+\frac{3\sqrt{2}-4\sqrt{3}}{6}=\frac{3\sqrt{2}}{1}+\frac{3\sqrt{2}-4\sqrt{3}}{6}-\sqrt{6}=\frac{18\sqrt{2}}{6}+\frac{3\sqrt{2}-4\sqrt{3}}{6}-\sqrt{6}=\\\\=\frac{18\sqrt{2}+3\sqrt{2}-4\sqrt{3}}{6}-\sqrt{6}=\frac{21\sqrt{2}-4\sqrt{3}}{6}-\sqrt{6}[/tex]