16 Вправа 16 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 318

16. Запишіть у вигляді степеня з основою 3:

1) ${\displaystyle 9^{-5}\cdot 3^7=}$${\displaystyle {\left(3^2\right)}^{-5}\cdot 3^7=}$${\displaystyle 3^{-10}\cdot 3^7=}$${\displaystyle 3^{-3}}$;

2) ${\displaystyle {27}^2\cdot 9^{-7}=}$${\displaystyle {\left(3^3\right)}^2\cdot {\left(3^2\right)}^{-7}=}$${\displaystyle 3^6\cdot 3^{-14}=}$${\displaystyle 3^{-8}}$;

3) ${\displaystyle {\left(9^{-2}\right)}^3\cdot {\left({27}^4\right)}^{-1}=}$${\displaystyle 9^{-6}\cdot {27}^{-4}=}$${\displaystyle {\left(3^2\right)}^{-6}\cdot {\left(3^3\right)}^{-4}=}$${\displaystyle 3^{-12}\cdot 3^{-12}=}$${\displaystyle 3^{-24}}$;

4) ${\displaystyle {\left({81}^2\right)}^{-9}\cdot {\left(9^{-2}\right)}^{-3}=}$${\displaystyle {81}^{-18}\cdot 9^6=}$${\displaystyle {\left(3^4\right)}^{-18}\cdot {\left(3^2\right)}^6=}$${\displaystyle 3^{-72}\cdot 3^{12}=}$${\displaystyle 3^{-60}}$;

5) ${\displaystyle {27}^{-20}:{\left(9^{-10}\right)}^3=}$${\displaystyle {\left(3^3\right)}^{-20}:9^{-30}=}$${\displaystyle 3^{-60}:{\left(3^2\right)}^{-30}=}$${\displaystyle 3^{-60}:3^{-60}=}$${\displaystyle 3^0}$;

6) ${\displaystyle {81}^{-5}:3^{20}=}$${\displaystyle {\left(3^4\right)}^{-5}:3^{20}=}$${\displaystyle 3^{-20}:3^{20}=}$${\displaystyle 3^{-40}}$.