Ты просишь найти значения выражений с разными степенями и числами: 2-9.(27)2

1. $2^{-9} \cdot (2^7)^2 = 2^{-9} \cdot 2^{14} = 2^{-9+14} = 2^5 = 32$ 2. $\frac{3^7}{81} = \frac{3^7}{3^4} = 3^{7-4} = 3^3 = 27$ 3. $\frac{(3^7)^{-2}}{3^{-16}} = \frac{3^{-14}}{3^{-16}} = 3^{-14-(-16)} = 3^{-14+16} = 3^2 = 9$ 4. $\frac{1}{2^{-11}} \cdot \frac{1}{2^7} = 2^{11} \cdot 2^{-7} = 2^{11-7} = 2^4 = 16$ 5. $\frac{9^{-6} \cdot 9^{15}}{9^7} = \frac{9^{-6+15}}{9^7} = \frac{9^9}{9^7} = 9^{9-7} = 9^2 = 81$ 6. $\frac{6^{12} \cdot 11^{10}}{66^{10}} = \frac{6^{12} \cdot 11^{10}}{(6 \cdot 11)^{10}} = \frac{6^{12} \cdot 11^{10}}{6^{10} \cdot 11^{10}} = 6^{12-10} \cdot 11^{10-10} = 6^2 \cdot 11^0 = 36 \cdot 1 = 36$ 7. $\frac{10^6}{2^5 \cdot 5^4} = \frac{(2 \cdot 5)^6}{2^5 \cdot 5^4} = \frac{2^6 \cdot 5^6}{2^5 \cdot 5^4} = 2^{6-5} \cdot 5^{6-4} = 2^1 \cdot 5^2 = 2 \cdot 25 = 50$ 8. $\frac{(3 \cdot 10)^8}{3^6 \cdot 10^7} = \frac{3^8 \cdot 10^8}{3^6 \cdot 10^7} = 3^{8-6} \cdot 10^{8-7} = 3^2 \cdot 10^1 = 9 \cdot 10 = 90$ 9. $a^{-13} \cdot (a^8)^2 = a^{-13} \cdot a^{16} = a^{-13+16} = a^3 = 3^3 = 27$ 10. $(a^4)^{-3} : a^{-15} = a^{-12} : a^{-15} = a^{-12 - (-15)} = a^{-12+15} = a^3 = 3^3 = 27$ 11. $\frac{(a^7)^2}{a^{12}} = \frac{a^{14}}{a^{12}} = a^{14-12} = a^2 = 5^2 = 25$ 12. $a^8 \cdot a^{17} : a^{20} = a^{8+17} : a^{20} = a^{25} : a^{20} = a^{25-20} = a^5 = 2^5 = 32$ 13. $\frac{a^9 \cdot a^{12}}{a^{18}} = \frac{a^{9+12}}{a^{18}} = \frac{a^{21}}{a^{18}} = a^{21-18} = a^3 = 4^3 = 64$ 14. $\frac{(a^9)^3 \cdot a^7}{a^{29}} = \frac{a^{27} \cdot a^7}{a^{29}} = \frac{a^{27+7}}{a^{29}} = \frac{a^{34}}{a^{29}} = a^{34-29} = a^5 = 2^5 = 32$ 15. $\frac{a^{23} \cdot (b^5)^4}{(a \cdot b)^{20}} = \frac{a^{23} \cdot b^{20}}{a^{20} \cdot b^{20}} = a^{23-20} \cdot b^{20-20} = a^3 \cdot b^0 = a^3 = 2^3 = 8$ **Ответы:** 1. 32 2. 27 3. 9 4. 16 5. 81 6. 36 7. 50 8. 90 9. 27 10. 27 11. 25 12. 32 13. 64 14. 32 15. 8