349 Вправа 349Алгебра 8 клас Тарасенкова 2025 · § 9 · стор. 97

349. Обчисліть:

1) ${\displaystyle {\left({\left({(-1)}^{-1}\right)}^{-1}\right)}^{-1}=}$${\displaystyle {(-1)}^{\left(-1\right)\left(-1\right)\left(-1\right)}=}$${\displaystyle {(-1)}^{-1}=}$${\displaystyle -1}$;

2) ${\displaystyle {\left({\left(2^{-1}\right)}^{-1}\right)}^{-1}=}$${\displaystyle 2^{\left(-1\right)\left(-1\right)\left(-1\right)}=}$${\displaystyle 2^{-1}=}$${\displaystyle \frac{1}{2}}$;

3) ${\displaystyle 4^{-4}\cdot {\left(\frac{1}{4}\right)}^{-5}=}$${\displaystyle 4^{-4}\cdot 4^5=}$${\displaystyle 4^{-4+5}=}$${\displaystyle 4}$;

4) ${\displaystyle {\left(2\frac{1}{2}\right)}^{-3}\cdot {\left(\frac{2}{5}\right)}^{-5}=}$${\displaystyle {\left(\frac{5}{2}\right)}^{-3}\cdot {\left(\frac{2}{5}\right)}^{-5}=}$${\displaystyle {\left(\frac{2}{5}\right)}^3\cdot {\left(\frac{5}{2}\right)}^5=}$${\displaystyle {\left(\frac{5}{2}\right)}^{-3+5}=}$${\displaystyle \frac{25}{4}=}$${\displaystyle 6\frac{1}{4}}$.