Ответ:

Решение:

  • 1) $$\frac{(5³)^{-4}}{5^{-11}} = \frac{5^{3 \times -4}}{5^{-11}} = \frac{5^{-12}}{5^{-11}} = 5^{-12 - (-11)} = 5^{-12+11} = 5^{-1} = \frac{1}{5}$$
  • 2) 5⁻⁷ · (5⁵)² = 5⁻⁷ · 5⁵*² = 5⁻⁷ · 5¹⁰ = 5⁻⁷⁺¹⁰ = 5³ = 125
  • 3) $$\frac{5^{-3} \cdot 5^{-9}}{5^{-11}} = \frac{5^{-3+(-9)}}{5^{-11}} = \frac{5^{-12}}{5^{-11}} = 5^{-12 - (-11)} = 5^{-12+11} = 5^{-1} = \frac{1}{5}$$
  • 4) (6 · 10²)³ · (13 · 10⁻⁵) = 6³ · (10²)³ · 13 · 10⁻⁵ = 216 · 10⁶ · 13 · 10⁻⁵ = (216 · 13) · 10⁶⁺⁽⁻⁵⁾ = 2808 · 10¹ = 28080
  • 5) $$\frac{7^{-3} \cdot 7^{13}}{7⁸} = \frac{7^{-3+13}}{7⁸} = \frac{7^{10}}{7⁸} = 7^{10-8} = 7² = 49$$
  • 6) $$\frac{(2⁴)^{-6}}{2⁻²⁷} = \frac{2^{4 \times -6}}{2⁻²⁷} = \frac{2^{-24}}{2⁻²⁷} = 2^{-24 - (-27)} = 2^{-24+27} = 2³ = 8$$
  • 7) 3⁻⁸ · (3⁶)² = 3⁻⁸ · 3⁶*² = 3⁻⁸ · 3¹² = 3⁻⁸⁺¹² = 3⁴ = 81
  • 8) $$\frac{4⁸ \cdot 11^{10}}{4⁴} = 4^{8-4} \cdot 11^{10} = 4⁴ \cdot 11^{10} = 256 \cdot 11^{10}$$

Ответ: 1) 1/5; 2) 125; 3) 1/5; 4) 28080; 5) 49; 6) 8; 7) 81; 8) 256 · 11¹⁰