Odpowiedź:
[tex]3.\\\\\sqrt{60}=\sqrt{4\cdot15}=\sqrt{4}\cdot\sqrt{15}=2\sqrt{15}\\\\Odp.C\\\\\\4.\\\\A.\ \ \sqrt{2}\cdot\sqrt{3}\cdot\sqrt{5}=\sqrt{2\cdot3\cdot5}=\sqrt{30}\\\\B.\ \ \frac{\sqrt{39}}{\sqrt{3}}=\sqrt{\frac{39}{3}}=\sqrt{13}\\\\C.\ \ \sqrt{40}\cdot\sqrt{\frac{1}{10}}=\sqrt{\not40^4\cdot\frac{1}{\not10_{1}}}=\sqrt{4}=2\\\\D.\ \ \sqrt{7}\cdot\sqrt{14}=\sqrt{7\cdot14}=\sqrt{98}=\sqrt{49\cdot2}=7\sqrt{2}\\\\Odp.C[/tex]
