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Решите задачу:

$(\sqrt{3 \frac{6}{7}} - \sqrt{1 \frac{5}{7}}): \sqrt{\frac{3}{28}} = (\sqrt{\frac{27}{7}} - \sqrt{\frac{12}{7}})\cdot\sqrt{\frac{28}{3}} = \sqrt{\frac{27}{7}}\cdot\sqrt{\frac{28}{3}} - \\ -\sqrt{\frac{12}{7}}\cdot\sqrt{\frac{28}{3}} = \sqrt{\frac{27\cdot28}{7\cdot3}} - \sqrt{\frac{12\cdot28}{7\cdot3}} = \sqrt{9\cdot4} - \sqrt{4\cdot4} = 6-4 = 2$

$\frac{\sqrt[9]{7}\cdot\sqrt[18]{7}}{\sqrt[13]{7}} = \frac{7^{\frac{1}{9}}\cdot7^{\frac{1}{18}}}{7^{\frac{1}{13}}}=7^{\frac{1}{9}+\frac{1}{18}-\frac{1}{13}}=7^{\frac{26+13-18}{234}}=7^{\frac{21}{234}}=7^{\frac{7}{78}}= \sqrt[78]{7^7}$