1201. Вычислите: a) 8-2.43 б) 9-6.275 в) 100 : 10-3 г) 125-4 : 25-5 д) 2-21.4-6 / 4-2.8-6 е) 4-2.8-6 / 2-22 ж) 3-10.98 / (-3)2 3) 5-5.2510 / 1253

Решение:

1. а) \( 8^{-2} \cdot 4^{3} = (2^{3})^{-2} \cdot (2^{2})^{3} = 2^{-6} \cdot 2^{6} = 2^{-6+6} = 2^{0} = 1 \)
2. б) \( 9^{-6} \cdot 27^{5} = (3^{2})^{-6} \cdot (3^{3})^{5} = 3^{-12} \cdot 3^{15} = 3^{-12+15} = 3^{3} = 27 \)
3. в) \( 10^{0} : 10^{-3} = 10^{0 - (-3)} = 10^{3} = 1000 \)
4. г) \( 125^{-4} : 25^{-5} = (5^{3})^{-4} : (5^{2})^{-5} = 5^{-12} : 5^{-10} = 5^{-12 - (-10)} = 5^{-12+10} = 5^{-2} = \frac{1}{25} \)
5. д) \( \frac{2^{-21} \cdot 4^{-6}}{4^{-2} \cdot 8^{-6}} = \frac{2^{-21} \cdot (2^{2})^{-6}}{(2^{2})^{-2} \cdot (2^{3})^{-6}} = \frac{2^{-21} \cdot 2^{-12}}{2^{-4} \cdot 2^{-18}} = \frac{2^{-21-12}}{2^{-4-18}} = \frac{2^{-33}}{2^{-22}} = 2^{-33 - (-22)} = 2^{-33+22} = 2^{-11} = \frac{1}{2048} \)
6. е) \( \frac{4^{-2} \cdot 8^{-6}}{2^{-22}} = \frac{(2^{2})^{-2} \cdot (2^{3})^{-6}}{2^{-22}} = \frac{2^{-4} \cdot 2^{-18}}{2^{-22}} = \frac{2^{-4-18}}{2^{-22}} = \frac{2^{-22}}{2^{-22}} = 1 \)
7. ж) \( \frac{3^{-10} \cdot 9^{8}}{(-3)^{2}} = \frac{3^{-10} \cdot (3^{2})^{8}}{3^{2}} = \frac{3^{-10} \cdot 3^{16}}{3^{2}} = \frac{3^{-10+16}}{3^{2}} = \frac{3^{6}}{3^{2}} = 3^{6-2} = 3^{4} = 81 \)
8. з) \( \frac{5^{-5} \cdot 25^{10}}{125^{3}} = \frac{5^{-5} \cdot (5^{2})^{10}}{(5^{3})^{3}} = \frac{5^{-5} \cdot 5^{20}}{5^{9}} = \frac{5^{-5+20}}{5^{9}} = \frac{5^{15}}{5^{9}} = 5^{15-9} = 5^{6} = 15625 \)

Ответ: а) 1; б) 27; в) 1000; г) 1/25; д) 1/2048; е) 1; ж) 81; з) 15625.