Вычисли и примени свойство корней.

1. Вычисляю: а) $\sqrt{25 \cdot 9} = \sqrt{25} \cdot \sqrt{9} = 5 \cdot 3 = 15$ б) $\sqrt{0{,}04 \cdot 81} = \sqrt{0{,}04} \cdot \sqrt{81} = 0{,}2 \cdot 9 = 1{,}8$ в) $\sqrt{36 \cdot 0{,}09 \cdot 25} = \sqrt{36} \cdot \sqrt{0{,}09} \cdot \sqrt{25} = 6 \cdot 0{,}3 \cdot 5 = 9$ г) $\sqrt{\frac{169}{25}} = \frac{\sqrt{169}}{\sqrt{25}} = \frac{13}{5} = 2{,}6$ д) $\sqrt{\frac{0{,}64 \cdot 49}{25}} = \frac{\sqrt{0{,}64} \cdot \sqrt{49}}{\sqrt{25}} = \frac{0{,}8 \cdot 7}{5} = \frac{5{,}6}{5} = 1{,}12$ е) $\sqrt{0{,}81 \cdot \frac{4}{25}} = \sqrt{0{,}81} \cdot \sqrt{\frac{4}{25}} = 0{,}9 \cdot \frac{2}{5} = 0{,}9 \cdot 0{,}4 = 0{,}36$ ж) $3(\sqrt{2})^2 - \frac{1}{3}(\sqrt{3})^2 = 3 \cdot 2 - \frac{1}{3} \cdot 3 = 6 - 1 = 5$ 2. Применяю свойство корней: а) $\sqrt{2} \cdot \sqrt{22} \cdot \sqrt{11} = \sqrt{2 \cdot 22 \cdot 11} = \sqrt{2 \cdot 2 \cdot 11 \cdot 11} = \sqrt{2^2 \cdot 11^2} = 2 \cdot 11 = 22$ в) $\sqrt{60} \cdot \sqrt{\frac{5}{3}} = \sqrt{60 \cdot \frac{5}{3}} = \sqrt{20 \cdot 5} = \sqrt{100} = 10$ б) $\frac{\sqrt{98}}{\sqrt{2}} - \frac{\sqrt{27}}{\sqrt{3}} = \sqrt{\frac{98}{2}} - \sqrt{\frac{27}{3}} = \sqrt{49} - \sqrt{9} = 7 - 3 = 4$ г) $\sqrt{\frac{11}{3}} : \sqrt{\frac{13}{25}} : \sqrt{\frac{13}{3}} = \sqrt{\frac{11}{3} : \frac{13}{25} : \frac{13}{3}} = \sqrt{\frac{11}{3} \cdot \frac{25}{13} \cdot \frac{3}{13}} = \sqrt{\frac{11 \cdot 25}{13 \cdot 13}} = \frac{5}{13} \sqrt{11}$