Д13 вероятность. 1) (455)2 80 2)5432 253 3) 532+ 518-752-3 4)(5-3)(53+3) 5) (842-5)² + 10 542 6) 8.8.15 8:12 71 2047 548.293 10)1 713 8) 3519-352-538 5+256 +5256

Ответ: Ниже приведены решения задач.

1. Задача 1: \[\frac{(4\sqrt{5})^2}{80}\]

   \[\frac{(4\sqrt{5})^2}{80} = \frac{16 \cdot 5}{80} = \frac{80}{80} = 1\]

2. Задача 2: \[\frac{\sqrt{432}}{2\sqrt{3}}\]

   \[\frac{\sqrt{432}}{2\sqrt{3}} = \frac{\sqrt{144 \cdot 3}}{2\sqrt{3}} = \frac{12\sqrt{3}}{2\sqrt{3}} = 6\]

3. Задача 3: \[\sqrt{32} + \sqrt{18} - \sqrt{50} - 3\]

   \[\sqrt{32} + \sqrt{18} - \sqrt{50} - 3 = \sqrt{16 \cdot 2} + \sqrt{9 \cdot 2} - \sqrt{25 \cdot 2} - 3 = 4\sqrt{2} + 3\sqrt{2} - 5\sqrt{2} - 3 = (4 + 3 - 5)\sqrt{2} - 3 = 2\sqrt{2} - 3\]

4. Задача 4: \[(\sqrt{13} - 3)(\sqrt{13} + 3)\]

   \[(\sqrt{13} - 3)(\sqrt{13} + 3) = (\sqrt{13})^2 - 3^2 = 13 - 9 = 4\]

5. Задача 5: \[(\sqrt{42} - 5)^2 + 10\sqrt{42}\]

   \[(\sqrt{42} - 5)^2 + 10\sqrt{42} = (42 - 10\sqrt{42} + 25) + 10\sqrt{42} = 42 + 25 = 67\]

6. Задача 6: \[\frac{8^{-6} \cdot 8^{-15}}{8^{-12}}\]

   \[\frac{8^{-6} \cdot 8^{-15}}{8^{-12}} = \frac{8^{-6-15}}{8^{-12}} = \frac{8^{-21}}{8^{-12}} = 8^{-21 - (-12)} = 8^{-21 + 12} = 8^{-9} = \frac{1}{8^9}\]

7. Задача 7: \[\frac{20^{47}}{5^{48} \cdot 2^{93}}\]

   \[\frac{20^{47}}{5^{48} \cdot 2^{93}} = \frac{(4 \cdot 5)^{47}}{5^{48} \cdot 2^{93}} = \frac{4^{47} \cdot 5^{47}}{5^{48} \cdot 2^{93}} = \frac{(2^2)^{47} \cdot 5^{47}}{5^{48} \cdot 2^{93}} = \frac{2^{94} \cdot 5^{47}}{5^{48} \cdot 2^{93}} = \frac{2}{5}\]

8. Задача 8: \[3\sqrt{19} \cdot 3\sqrt{2} \cdot \sqrt{38}\]

   \[3\sqrt{19} \cdot 3\sqrt{2} \cdot \sqrt{38} = 9\sqrt{19} \cdot \sqrt{2 \cdot 38} = 9\sqrt{19} \cdot \sqrt{76} = 9\sqrt{19} \cdot \sqrt{4 \cdot 19} = 9\sqrt{19} \cdot 2\sqrt{19} = 18 \cdot 19 = 342\]

9. Задача 9: \[\frac{1}{7^{-14}} \cdot \frac{1}{7^{13}}\]

   \[\frac{1}{7^{-14}} \cdot \frac{1}{7^{13}} = 7^{14} \cdot \frac{1}{7^{13}} = \frac{7^{14}}{7^{13}} = 7^{14-13} = 7\]

10. Задача 10: \[\frac{1}{5+2\sqrt{6}} + \frac{1}{5-2\sqrt{6}}\]

    \[\frac{1}{5+2\sqrt{6}} + \frac{1}{5-2\sqrt{6}} = \frac{(5-2\sqrt{6}) + (5+2\sqrt{6})}{(5+2\sqrt{6})(5-2\sqrt{6})} = \frac{10}{25 - (4 \cdot 6)} = \frac{10}{25 - 24} = \frac{10}{1} = 10\]

Ответ: Задача 1: 1, Задача 2: 6, Задача 3: 2\sqrt{2} - 3, Задача 4: 4, Задача 5: 67, Задача 6: \frac{1}{8^9}, Задача 7: \frac{2}{5}, Задача 8: 342, Задача 9: 7, Задача 10: 10