348 Вправа 348Алгебра 8 клас Тарасенкова 2025 · § 9 · стор. 97

348. Обчисліть:

Степінь степеня: показники перемножуємо.

1) ${\displaystyle {\left({\left({\left(5^{-1}\right)}^{-1}\right)}^{-1}\right)}^{-1}=}$${\displaystyle 5^{\left(-1\right)\left(-1\right)\left(-1\right)\left(-1\right)}=}$${\displaystyle 5^1=}$${\displaystyle 5}$;

2) ${\displaystyle {\left({\left(9^{-2}\right)}^{-1}\right)}^0=}$${\displaystyle 9^{\left(-2\right)\left(-1\right)\cdot 0}=}$${\displaystyle 9^0=}$${\displaystyle 1}$;

3) ${\displaystyle {\left({\left({\left(3^{-1}\right)}^2\right)}^{-1}\right)}^2=}$${\displaystyle 3^{\left(-1\right)\cdot 2\cdot \left(-1\right)\cdot 2}=}$${\displaystyle 3^4=}$${\displaystyle 81}$;

4) ${\displaystyle {\left({\left({\left({0,5}^{-1}\right)}^2\right)}^{-1}\right)}^{-2}=}$${\displaystyle {0,5}^{\left(-1\right)\cdot 2\cdot \left(-1\right)\cdot \left(-2\right)}=}$${\displaystyle {0,5}^{-4}=}$${\displaystyle 2^4=}$${\displaystyle 16}$;

5) ${\displaystyle 3^{-1}\cdot {\left(\frac{1}{3}\right)}^3=}$${\displaystyle \frac{1}{3}\cdot \frac{1}{27}=}$${\displaystyle \frac{1}{81}}$;

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

7) ${\displaystyle {\left(\frac{1}{7}\right)}^{-6}:7^7\cdot 2^{-1}=}$${\displaystyle 7^6:7^7\cdot \frac{1}{2}=}$${\displaystyle 7^{-1}\cdot \frac{1}{2}=}$${\displaystyle \frac{1}{14}}$;

8) ${\displaystyle -{(1,2)}^4:{\left(\frac{5}{6}\right)}^5=}$${\displaystyle -{\left(\frac{6}{5}\right)}^4\cdot {\left(\frac{6}{5}\right)}^5=}$${\displaystyle -{\left(\frac{6}{5}\right)}^9=}$${\displaystyle -{1,2}^9}$;

9) ${\displaystyle 3\cdot {0,75}^{-1}:{\left(\frac{1}{2}\right)}^4=}$${\displaystyle 3\cdot \frac{4}{3}:\frac{1}{16}=}$${\displaystyle 4\cdot 16=}$${\displaystyle 64}$;

10) ${\displaystyle {28}^0\cdot {0,5}^{-3}+{0,2}^3\cdot {0,001}^{-1}=}$${\displaystyle 1\cdot 2^3+0,008\cdot 1000=}$${\displaystyle 8+8=}$${\displaystyle 16}$.