Вычисли значения выражений с квадратными корнями.

1) $\frac{\sqrt{72}}{\sqrt{2}} = \sqrt{\frac{72}{2}} = \sqrt{36} = 6$ 2) $\frac{\sqrt{80}}{\sqrt{5}} = \sqrt{\frac{80}{5}} = \sqrt{16} = 4$ 3) $\frac{\sqrt{75}}{\sqrt{3}} = \sqrt{\frac{75}{3}} = \sqrt{25} = 5$ 4) $\sqrt{(-5)^2} = \sqrt{25} = 5$ 5) $\sqrt{(-17)^2} = \sqrt{289} = 17$ 6) $\sqrt{(-8)^2} = \sqrt{64} = 8$ 7) $\sqrt{56 \cdot 40 \cdot 35} = \sqrt{(7 \cdot 8) \cdot (5 \cdot 8) \cdot (5 \cdot 7)} = \sqrt{7^2 \cdot 8^2 \cdot 5^2} = 7 \cdot 8 \cdot 5 = 280$ 8) $\sqrt{66 \cdot 110 \cdot 15} = \sqrt{(6 \cdot 11) \cdot (10 \cdot 11) \cdot (3 \cdot 5)} = \sqrt{(2 \cdot 3 \cdot 11) \cdot (2 \cdot 5 \cdot 11) \cdot (3 \cdot 5)} = \sqrt{2^2 \cdot 3^2 \cdot 5^2 \cdot 11^2} = 2 \cdot 3 \cdot 5 \cdot 11 = 330$ 9) $\sqrt{48 \cdot 80 \cdot 15} = \sqrt{(16 \cdot 3) \cdot (16 \cdot 5) \cdot (3 \cdot 5)} = \sqrt{16^2 \cdot 3^2 \cdot 5^2} = 16 \cdot 3 \cdot 5 = 240$ 10) $\sqrt{5^6} = 5^{6/2} = 5^3 = 125$ 11) $\sqrt{9^3} = 9^{3/2} = (3^2)^{3/2} = 3^3 = 27$ 12) $\sqrt{6^4} = 6^{4/2} = 6^2 = 36$ 13) $\sqrt{10 \cdot 7^2 \cdot 10 \cdot 2^6} = \sqrt{10^2 \cdot 7^2 \cdot 2^6} = 10 \cdot 7 \cdot 2^{6/2} = 10 \cdot 7 \cdot 2^3 = 10 \cdot 7 \cdot 8 = 560$ 14) $\sqrt{8 \cdot 2^{12} \cdot 8 \cdot 5^4} = \sqrt{8^2 \cdot 2^{12} \cdot 5^4} = 8 \cdot 2^{12/2} \cdot 5^{4/2} = 8 \cdot 2^6 \cdot 5^2 = 8 \cdot 64 \cdot 25 = 12800$ 15) $\sqrt{11 \cdot 3^2 \cdot 11 \cdot 4^4} = \sqrt{11^2 \cdot 3^2 \cdot 4^4} = 11 \cdot 3 \cdot 4^{4/2} = 11 \cdot 3 \cdot 4^2 = 11 \cdot 3 \cdot 16 = 528$ 16) $\frac{72}{(2\sqrt{3})^2} = \frac{72}{2^2 \cdot (\sqrt{3})^2} = \frac{72}{4 \cdot 3} = \frac{72}{12} = 6$ 17) $\frac{90}{(3\sqrt{5})^2} = \frac{90}{3^2 \cdot (\sqrt{5})^2} = \frac{90}{9 \cdot 5} = \frac{90}{45} = 2$ 18) $\frac{(2\sqrt{5})^2}{160} = \frac{2^2 \cdot (\sqrt{5})^2}{160} = \frac{4 \cdot 5}{160} = \frac{20}{160} = \frac{1}{8}$ 19) $(\sqrt{19} - \sqrt{2})(\sqrt{19} + \sqrt{2}) = (\sqrt{19})^2 - (\sqrt{2})^2 = 19 - 2 = 17$ 20) $(\sqrt{18} - \sqrt{6})(\sqrt{6} + \sqrt{18}) = (\sqrt{18} - \sqrt{6})(\sqrt{18} + \sqrt{6}) = (\sqrt{18})^2 - (\sqrt{6})^2 = 18 - 6 = 12$ 21) $(\sqrt{14} + \sqrt{15})(\sqrt{15} - \sqrt{14}) = (\sqrt{15} + \sqrt{14})(\sqrt{15} - \sqrt{14}) = (\sqrt{15})^2 - (\sqrt{14})^2 = 15 - 14 = 1$ 22) $\frac{\sqrt{51} \cdot \sqrt{12}}{\sqrt{17}} = \frac{\sqrt{51 \cdot 12}}{\sqrt{17}} = \sqrt{\frac{51 \cdot 12}{17}} = \sqrt{\frac{(3 \cdot 17) \cdot 12}{17}} = \sqrt{3 \cdot 12} = \sqrt{36} = 6$ 23) $\frac{\sqrt{21} \cdot \sqrt{14}}{\sqrt{6}} = \frac{\sqrt{21 \cdot 14}}{\sqrt{6}} = \sqrt{\frac{21 \cdot 14}{6}} = \sqrt{\frac{(3 \cdot 7) \cdot (2 \cdot 7)}{2 \cdot 3}} = \sqrt{7^2} = 7$ 24) $\frac{\sqrt{65} \cdot \sqrt{13}}{\sqrt{5}} = \frac{\sqrt{65 \cdot 13}}{\sqrt{5}} = \sqrt{\frac{65 \cdot 13}{5}} = \sqrt{\frac{(5 \cdot 13) \cdot 13}{5}} = \sqrt{13^2} = 13$ 25) $4\sqrt{17} \cdot 5\sqrt{2} \cdot \sqrt{34} = 4 \cdot 5 \cdot \sqrt{17 \cdot 2 \cdot 34} = 20 \cdot \sqrt{17 \cdot 2 \cdot (2 \cdot 17)} = 20 \cdot \sqrt{17^2 \cdot 2^2} = 20 \cdot 17 \cdot 2 = 680$ 26) $9\sqrt{7} \cdot 2\sqrt{2} \cdot \sqrt{14} = 9 \cdot 2 \cdot \sqrt{7 \cdot 2 \cdot 14} = 18 \cdot \sqrt{7 \cdot 2 \cdot (2 \cdot 7)} = 18 \cdot \sqrt{7^2 \cdot 2^2} = 18 \cdot 7 \cdot 2 = 252$ 27) $10\sqrt{7} \cdot 2\sqrt{6} \cdot \sqrt{42} = 10 \cdot 2 \cdot \sqrt{7 \cdot 6 \cdot 42} = 20 \cdot \sqrt{7 \cdot 6 \cdot (6 \cdot 7)} = 20 \cdot \sqrt{7^2 \cdot 6^2} = 20 \cdot 7 \cdot 6 = 840$