338° Вправа 338Алгебра 8 клас Тарасенкова 2025 · § 9 · стор. 95

338. Запишіть як степінь з основою 30:

1) ${\displaystyle 3^{-5}\cdot {10}^{-5}=}$${\displaystyle {(3\cdot 10)}^{-5}=}$${\displaystyle {30}^{-5}}$;

2) ${\displaystyle 6^{-9}\cdot 5^{-9}=}$${\displaystyle {(6\cdot 5)}^{-9}=}$${\displaystyle {30}^{-9}}$;

3) ${\displaystyle 2^{-6}\cdot {15}^{-6}=}$${\displaystyle {(2\cdot 15)}^{-6}=}$${\displaystyle {30}^{-6}}$;

4) ${\displaystyle 2^{-3}\cdot 3^{-3}\cdot 5^{-3}=}$${\displaystyle {(2\cdot 3\cdot 5)}^{-3}=}$${\displaystyle {30}^{-3}}$;

5) ${\displaystyle {60}^{-1}:2^{-1}=}$${\displaystyle {(60:2)}^{-1}=}$${\displaystyle {30}^{-1}}$;

6) ${\displaystyle {900}^{-2}:{30}^{-2}=}$${\displaystyle {(900:30)}^{-2}=}$${\displaystyle {30}^{-2}}$;

7) ${\displaystyle {15}^{-4}\cdot 4^{-4}:2^{-4}=}$${\displaystyle {(15\cdot 4:2)}^{-4}=}$${\displaystyle {30}^{-4}}$;

8) ${\displaystyle {40}^{-11}:4^{-11}\cdot 3^{-11}=}$${\displaystyle {(40:4\cdot 3)}^{-11}=}$${\displaystyle {30}^{-11}}$;

9) ${\displaystyle {120}^{-3}:4^{-3}\cdot {17}^0=}$${\displaystyle {(120:4)}^{-3}\cdot 1=}$${\displaystyle {30}^{-3}}$.