вправа 238 гдз 7 клас алгебра Мерзляк Полонський

Відповідь: \begin{equation}1)\left ( 1\frac{1}{6} \right )^{9}\cdot \left ( \frac{6}{7} \right )^{10}=\end{equation} \begin{equation}=\left ( \frac{7}{6} \right )^{9}\cdot \left ( \frac{6}{7} \right )^{9}\cdot \frac{6}{7}=\end{equation} \begin{equation}=\left ( \frac{7}{6}\cdot \frac{6}{7} \right )^{9}\cdot \frac{6}{7}=\end{equation} \begin{equation}=1^{9}\cdot \frac{6}{7}=\frac{6}{7};\end{equation} \begin{equation}2)5^{14}\cdot 0,2^{12}=\end{equation} \begin{equation}=5^{2}\cdot 5^{12}\cdot 0,2^{12}=\end{equation} \begin{equation}=5^{2}\cdot \left ( 5\cdot 0,2 \right )^{12}=\end{equation} \begin{equation}=25\cdot 1^{12}=25;\end{equation} \begin{equation}3)\left ( -1\frac{1}{3} \right )^{5}\cdot \left ( \frac{3}{4} \right )^{8}=\end{equation} \begin{equation}=\left ( -1 \right )^{5}\cdot \left ( \frac{4}{3} \right )^{5}\cdot \left ( \frac{3}{4} \right )^{8}=\end{equation} \begin{equation}=-\left ( \frac{4}{3} \right )^{5}\cdot \left ( \frac{3}{4} \right )^{5}\cdot \left ( \frac{3}{4} \right )^{3}=\end{equation} \begin{equation}=-\left ( \frac{4}{3}\cdot \frac{3}{4} \right )^{5}\cdot \left ( \frac{3}{4} \right )^{3}=-\frac{27}{64}.\end{equation}