Odpowiedź:
[tex]Zad.2\\\\\sqrt{108}-4\sqrt{75}+5\sqrt{12}=\sqrt{36\cdot3}-4\sqrt{25\cdot3}+5\sqrt{4\cdot3}=\\\\=\sqrt{36}\cdot\sqrt{3}-4\cdot\sqrt{25}\cdot\sqrt{3}+5\cdot\sqrt{4}\cdot\sqrt{3}=6\sqrt{3}-4\cdot5\sqrt{3}+5\cdot2\sqrt{3}=\\\\=6\sqrt{3}-20\sqrt{3}+10\sqrt{3}=-4\sqrt{3}[/tex]
[tex]Zad.3\\\\3\sqrt{2}+\frac{5}{7\sqrt{2}}=3\sqrt{2}+\frac{5}{7\sqrt{2}}\cdot\frac{\sqrt{2} }{\sqrt{2}}=3\sqrt{2}+\frac{5\sqrt{2} }{7\cdot2}=3\sqrt{2}+\frac{5\sqrt{2}}{14}=\frac{3\sqrt{2}}{1}+\frac{5\sqrt{2}}{14}=\\\\=\frac{14\cdot3\sqrt{2}}{14}+\frac{5\sqrt{2}}{14}= \frac{42\sqrt{2}}{14}+\frac{5\sqrt{2}}{14}=\frac{47\sqrt{2}}{14}[/tex]
