a) \( a^5 \cdot a^3 = a^{5+3} = a^8 \)
б) \( a^{-6} : a^{-9} = a^{-6 - (-9)} = a^{-6 + 9} = a^3 \)
в) \( (a^5)^{-3} \cdot a^{18} = a^{5 \cdot (-3)} \cdot a^{18} = a^{-15} \cdot a^{18} = a^{-15 + 18} = a^3 \)
а) \( 2\,450\,000 = 2.45 \cdot 10^6 \)
б) \( 0.000019 = 1.9 \cdot 10^{-5} \)
а) \( 5^{-4} \cdot 5^2 = 5^{-4 + 2} = 5^{-2} = \frac{1}{5^2} = \frac{1}{25} = 0.04 \)
б) \( 12^{-3} : 12^{-4} = 12^{-3 - (-4)} = 12^{-3 + 4} = 12^1 = 12 \)
в) \( (3^{-1})^{-3} = 3^{(-1) \cdot (-3)} = 3^3 = 27 \)
а) \( (\frac{1}{6}x^{-4}y^3)^{-1} = (\frac{1}{6})^{-1} \cdot (x^{-4})^{-1} \cdot (y^3)^{-1} = 6x^4y^{-3} = \frac{6x^4}{y^3} \)
б) \( (\frac{2b}{3a})^{-2} \cdot 12ab^2 = (\frac{3a}{2b})^2 \cdot 12ab^2 = \frac{9a^2}{4b^2} \cdot 12ab^2 = \frac{9 \cdot 12 \cdot a^2 \cdot a \cdot b^2}{4b^2} = 27a^3 \)
в) \( 0.2a^8b^{-10} \cdot 1.7a^{-6}b^{12} = 0.2 \cdot 1.7 \cdot a^{8 + (-6)} \cdot b^{-10 + 12} = 0.34a^2b^2 \)
а) \( 6^{-2} + (\frac{3}{4})^{-1} = \frac{1}{6^2} + \frac{4}{3} = \frac{1}{36} + \frac{4}{3} = \frac{1}{36} + \frac{48}{36} = \frac{49}{36} = 1\frac{13}{36} \)
б) \( \frac{8^{-4} \cdot 8^{-9}}{8^{-15}} = \frac{8^{-4 + (-9)}}{8^{-15}} = \frac{8^{-13}}{8^{-15}} = 8^{-13 - (-15)} = 8^{-13 + 15} = 8^2 = 64 \)
а) \( (27 \cdot 3^{-6})^2 \cdot (9^{-1})^{-2} = (3^3 \cdot 3^{-6})^2 \cdot (3^{-2})^{-2} = (3^{3 - 6})^2 \cdot 3^{(-2) \cdot (-2)} = (3^{-3})^2 \cdot 3^4 = 3^{-6} \cdot 3^4 = 3^{-6 + 4} = 3^{-2} = \frac{1}{3^2} = \frac{1}{9} \)
б) \( \frac{(-64)^{-4} \cdot 8^3}{16^{-3}} = \frac{((-2)^6)^{-4} \cdot (2^3)^3}{(2^4)^{-3}} = \frac{(-2)^{-24} \cdot 2^9}{2^{-12}} = \frac{2^{-24} \cdot 2^9}{2^{-12}} = 2^{-24 + 9 - (-12)} = 2^{-24 + 9 + 12} = 2^{-3} = \frac{1}{2^3} = \frac{1}{8} \)
Ответ: См. решение выше