Вычислите: 2^-4; (-7)^-2; (3/7)^-1; (3,91)^0.

Ответ:

\[2^{- 4} = \frac{1}{2^{4}} = \frac{1}{16};\text{\ \ \ \ }\]

\[( - 7)^{- 2} = \frac{1}{( - 7)^{2}\ } = \frac{1}{49};\text{\ \ }\]

\[\left( \frac{3}{7} \right)^{- 1} = \frac{7}{3};\ \ \ \ \ \ \ \ \]

\[(3,91)^{0} = 1.\]