3ana HHe 1 BbIpaxKeHHS: 1 ((8^3)^-7)/(8^-23) 7. (2^-7cdot 2^-8)/(2^-16) 13 (1)/(5^-8)cdot (1)/(5^6) ((3^7)^-2)/(3^-16) (9^-5cdot 9^-8)/(9^-15) (14) (1)/(7^-14)cdot (1)/(7^13) 3 ((2^9)^-3)/(2^-29) 9 (3^-4cdot 3^-8)/(3^-14) 15 (1)/(2^-19)cdot (1)/(2^16) ((5^2)^-8)/(5^-18) 10 (7^-3cdot 7^-8)/(7^-13) 16 (1)/(8^-7)cdot (1)/(8^6) 5 ((7^7)^-3)/(7^-23) 11 (11^-5cdot 11^-13)/(11^-19) 17 (1)/(3^-10)cdot (1)/(3^8) ((6^2)^-9)/(6^-20) ( 12 (5^-3cdot 5^-9)/(5^-14) 18 (1)/(4^-10):(1)/(4^9)

1. $\frac {(8^{3})^{-7}}{8^{-23}} = 8^{-21} \cdot 8^{23} = 8^{2} = 64$<br /><br />7. $\frac {2^{-7}\cdot 2^{-8}}{2^{-16}} = 2^{-15} \cdot 2^{16} = 2^{1} = 2$<br /><br />13. $\frac {1}{5^{-8}}\cdot \frac {1}{5^{6}} = 5^{8} \cdot 5^{-6} = 5^{2} = 25$<br /><br />$\frac {(3^{7})^{-2}}{3^{-16}} = 3^{-14} \cdot 3^{16} = 3^{2} = 9$<br /><br />$\frac {9^{-5}\cdot 9^{-8}}{9^{-15}} = 9^{-13} \cdot 9^{15} = 9^{2} = 81$<br /><br />(14) $\frac {1}{7^{-14}}\cdot \frac {1}{7^{13}} = 7^{14} \cdot 7^{-13} = 7^{1} = 7$<br /><br />3. $\frac {(2^{9})^{-3}}{2^{-29}} = 2^{-27} \cdot 2^{29} = 2^{2} = 4$<br /><br />9. $\frac {3^{-4}\cdot 3^{-8}}{3^{-14}} = 3^{-12} \cdot 3^{14} = 3^{2} = 9$<br /><br />15. $\frac {1}{2^{-19}}\cdot \frac {1}{2^{16}} = 2^{19} \cdot 2^{16} = 2^{35}$<br /><br />$\frac {(5^{2})^{-8}}{5^{-18}} = 5^{-16} \cdot 5^{18} = 5^{2} = 25$<br /><br />10. $\frac {7^{-3}\cdot 7^{-8}}{7^{-13}} = 7^{-11} \cdot 7^{13} = 7^{2} = 49$<br /><br />16. $\frac {1}{8^{-7}}\cdot \frac {1}{8^{6}} = 8^{7} \cdot 8^{-6} = 8^{1} = 8$<br /><br />5. $\frac {(7^{7})^{-3}}{7^{-23}} = 7^{-21} \cdot 7^{23} = 7^{2} = 49$<br /><br />11. $\frac {11^{-5}\cdot 11^{-13}}{11^{-19}} = 11^{-18} \cdot 11^{19} = 11^{1} = 11$<br /><br />17. $\frac {1}{3^{-10}}\cdot \frac {1}{3^{8}} = 3^{10} \cdot 3^{-8} = 3^{2} = 9$<br /><br />$\frac {(6^{2})^{-9}}{6^{-20}} = 6^{-18} \cdot 6^{20} = 6^{2} = 36$<br /><br />(12) $\frac {5^{-3}\cdot 5^{-9}}{5^{-14}} = 5^{-12} \cdot 5^{14} = 5^{2} = 25$<br /><br />18. $\frac {1}{4^{-10}}:\frac {1}{4^{9}} = 4^{10} \cdot 4^{9} = 4^{19}$