Вычисли значение выражения (3√6)²/18

Конечно, давай вычислим значения этих выражений! 1) $\frac{(3\sqrt{6})^2}{18} = \frac{9 \cdot 6}{18} = \frac{54}{18} = 3$ 2) $\frac{(4\sqrt{5})^2}{80} = \frac{16 \cdot 5}{80} = \frac{80}{80} = 1$ 3) $\frac{14}{(2\sqrt{7})^2} = \frac{14}{4 \cdot 7} = \frac{14}{28} = \frac{1}{2} = 0.5$ 4) $\frac{147}{(5\sqrt{21})^2} = \frac{147}{25 \cdot 21} = \frac{147}{525} = \frac{7}{25} = 0.28$ 5) $\frac{\sqrt{200}}{4\sqrt{2}} = \frac{\sqrt{100 \cdot 2}}{4\sqrt{2}} = \frac{10\sqrt{2}}{4\sqrt{2}} = \frac{10}{4} = 2.5$ 6) $\frac{\sqrt{432}}{2\sqrt{3}} = \frac{\sqrt{144 \cdot 3}}{2\sqrt{3}} = \frac{12\sqrt{3}}{2\sqrt{3}} = \frac{12}{2} = 6$ 7) $\frac{\sqrt{200}}{\sqrt{8}} = \sqrt{\frac{200}{8}} = \sqrt{25} = 5$ 8) $\frac{\sqrt{175}}{\sqrt{7}} = \sqrt{\frac{175}{7}} = \sqrt{25} = 5$ 9) $\sqrt{45 \cdot 27 \cdot 60} = \sqrt{9 \cdot 5 \cdot 9 \cdot 3 \cdot 4 \cdot 15} = \sqrt{9 \cdot 9 \cdot 4 \cdot 5 \cdot 3 \cdot 15} = 3 \cdot 3 \cdot 2 \cdot \sqrt{5 \cdot 45} = 18 \cdot \sqrt{225} = 18 \cdot 15 = 270$ 10) $\sqrt{18 \cdot 12 \cdot 24} = \sqrt{9 \cdot 2 \cdot 4 \cdot 3 \cdot 4 \cdot 6} = \sqrt{9 \cdot 4 \cdot 4 \cdot 2 \cdot 3 \cdot 6} = 3 \cdot 2 \cdot 2 \cdot \sqrt{36} = 12 \cdot 6 = 72$ 11) $\sqrt{8} \cdot \sqrt{2} + 3 = \sqrt{16} + 3 = 4 + 3 = 7$ 12) $\sqrt{3} \cdot \sqrt{12} - 5 = \sqrt{36} - 5 = 6 - 5 = 1$ 13) $\sqrt{16^4} = 16^2 = 256$ 14) $\sqrt{5^6} = 5^3 = 125$ 15) $\sqrt{72} + \sqrt{8} - 8\sqrt{2} + 2 = \sqrt{36 \cdot 2} + \sqrt{4 \cdot 2} - 8\sqrt{2} + 2 = 6\sqrt{2} + 2\sqrt{2} - 8\sqrt{2} + 2 = 0\sqrt{2} + 2 = 2$ 16) $\sqrt{32} + \sqrt{18} - 7\sqrt{2} - 3 = \sqrt{16 \cdot 2} + \sqrt{9 \cdot 2} - 7\sqrt{2} - 3 = 4\sqrt{2} + 3\sqrt{2} - 7\sqrt{2} - 3 = 0\sqrt{2} - 3 = -3$ 17) $(\sqrt{10} - 6)(\sqrt{10} + 6) = (\sqrt{10})^2 - 6^2 = 10 - 36 = -26$ 18) $(\sqrt{13} - 3)(\sqrt{13} + 3) = (\sqrt{13})^2 - 3^2 = 13 - 9 = 4$ 19) $\sqrt{54} - \sqrt{24} - \sqrt{6} + 12 = \sqrt{9 \cdot 6} - \sqrt{4 \cdot 6} - \sqrt{6} + 12 = 3\sqrt{6} - 2\sqrt{6} - \sqrt{6} + 12 = 0\sqrt{6} + 12 = 12$ 20) $\sqrt{72} - \sqrt{8} - 4\sqrt{2} - 13 = \sqrt{36 \cdot 2} - \sqrt{4 \cdot 2} - 4\sqrt{2} - 13 = 6\sqrt{2} - 2\sqrt{2} - 4\sqrt{2} - 13 = 0\sqrt{2} - 13 = -13$ 21) $(\sqrt{62} + 3)^2 - 6\sqrt{62} = 62 + 6\sqrt{62} + 9 - 6\sqrt{62} = 71$ 22) $(\sqrt{42} - 5)^2 + 10\sqrt{42} = 42 - 10\sqrt{42} + 25 + 10\sqrt{42} = 67$ 23) $\frac{(6^5)^{-6}}{6^{-32}} = \frac{6^{-30}}{6^{-32}} = 6^{-30 - (-32)} = 6^{-30 + 32} = 6^2 = 36$ 24) $\frac{(8^4)^{-5}}{8^{-19}} = \frac{8^{-20}}{8^{-19}} = 8^{-20 - (-19)} = 8^{-20 + 19} = 8^{-1} = \frac{1}{8} = 0.125$ 25) $3^{-11} \cdot (3^5)^3 = 3^{-11} \cdot 3^{15} = 3^{-11 + 15} = 3^4 = 81$ 26) $9^{-5} \cdot (9^3)^2 = 9^{-5} \cdot 9^6 = 9^{-5 + 6} = 9^1 = 9$ 27) $\frac{4^{-2} \cdot 4^{-6}}{4^{-10}} = \frac{4^{-8}}{4^{-10}} = 4^{-8 - (-10)} = 4^{-8 + 10} = 4^2 = 16$ 28) $\frac{8^{-6} \cdot 8^{-5}}{8^{-12}} = \frac{8^{-11}}{8^{-12}} = 8^{-11 - (-12)} = 8^{-11 + 12} = 8^1 = 8$ 29) $\frac{4^{15}}{8^9} = \frac{(2^2)^{15}}{(2^3)^9} = \frac{2^{30}}{2^{27}} = 2^{30 - 27} = 2^3 = 8$ 30) $\frac{36^{11}}{216^7} = \frac{(6^2)^{11}}{(6^3)^7} = \frac{6^{22}}{6^{21}} = 6^{22 - 21} = 6^1 = 6$ 31) $\frac{3^{17} \cdot 6^{16}}{18^{15}} = \frac{3^{17} \cdot (3 \cdot 2)^{16}}{(3^2 \cdot 2)^{15}} = \frac{3^{17} \cdot 3^{16} \cdot 2^{16}}{3^{30} \cdot 2^{15}} = \frac{3^{33} \cdot 2^{16}}{3^{30} \cdot 2^{15}} = 3^{33 - 30} \cdot 2^{16 - 15} = 3^3 \cdot 2^1 = 27 \cdot 2 = 54$ 32) $\frac{20^{47}}{5^{48} \cdot 2^{93}} = \frac{(4 \cdot 5)^{47}}{5^{48} \cdot 2^{93}} = \frac{(2^2 \cdot 5)^{47}}{5^{48} \cdot 2^{93}} = \frac{2^{94} \cdot 5^{47}}{5^{48} \cdot 2^{93}} = 2^{94 - 93} \cdot 5^{47 - 48} = 2^1 \cdot 5^{-1} = 2 \cdot \frac{1}{5} = \frac{2}{5} = 0.4$ Надеюсь, теперь тебе стало понятнее!