Задание 14, стр 29

Question

Вычислите. 
$$a)8^{-2}\cdot 4^3\quad b)9^{-2}\cdot 27^{2}\quad c)\frac{16^{-12}}{4^{-21}\cdot 8^{-3}}\quad d) \frac{3^{-12}\cdot 9^4}{27^{-2}}.$$

Gimli · Accepted Answer

\begin{enumerate}
\item $$8^{-2}\cdot 4^3 = \frac{1}{8^2}\cdot 4^3 = \frac{1}{4^2\cdot 2^2}\cdot 4^3 = \frac{1}{2^2}\cdot 4 = 1.$$
\item $$9^{-2}\cdot 27^2 = \frac{1}{9^2}\cdot 9^2\cdot 3^2 = 3^2=9.$$
\item $$\frac{\frac{1}{16^{12}}}{\frac{1}{4^{21}}\cdot \frac{1}{8^3}} = \frac{4^{21}\cdot 8^3}{16^{12}}=\frac{4^{21}\cdot 4^3\cdot 2^3}{4^{12}\cdot 4^{12}} = \frac{4^{24}\cdot 2^3}{4^{24}} = 2^3=8.$$
\item $$\frac{3^{-12}\cdot 9^4}{27^{-2}} = \frac{\frac{1}{3^{12}}\cdot 9^4}{\frac{1}{27^2}} = \frac{9^4\cdot 27^2}{3^{12}} = \frac{(3^2)^4\cdot (3^3)^2}{3^{12}} = \frac{3^8\cdot 3^6}{3^{12}} = \frac{3^14}{3^12} = 3^{14-12} = 3^2 = 9.$$
\end{enumerate}

Задание 14, стр 29

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Задание 14, стр 29

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