318° Вправа 318Алгебра 8 клас Тарасенкова 2025 · § 9 · стор. 93

318. Обчисліть:

${\displaystyle a^n\cdot a^m=}$${\displaystyle a^{n+m}}$ — показники додаємо.

1) ${\displaystyle 2^{-4}\cdot 2^{-2}=}$${\displaystyle 2^{-4+\left(-2\right)}=}$${\displaystyle 2^{-6}=}$${\displaystyle \frac{1}{64}}$;

2) ${\displaystyle 2^4\cdot 2^{-2}=}$${\displaystyle 2^{4-2}=}$${\displaystyle 4}$;

3) ${\displaystyle 2^{-4}\cdot 2^2=}$${\displaystyle 2^{-4+2}=}$${\displaystyle 2^{-2}=}$${\displaystyle \frac{1}{4}}$;

4) ${\displaystyle 2^4\cdot 2^2=}$${\displaystyle 2^{4+2}=}$${\displaystyle 64}$;

5) ${\displaystyle 2^8\cdot 2^{-3}=}$${\displaystyle 2^{8-3}=}$${\displaystyle 32}$;

6) ${\displaystyle 2^{-7}\cdot 2^3=}$${\displaystyle 2^{-7+3}=}$${\displaystyle 2^{-4}=}$${\displaystyle \frac{1}{16}}$;

7) ${\displaystyle 2^5\cdot 2^{-11}\cdot 2^0=}$${\displaystyle 2^{5-11+0}=}$${\displaystyle 2^{-6}=}$${\displaystyle \frac{1}{64}}$;

8) ${\displaystyle 2\cdot 2^{13}\cdot 2^{-16}=}$${\displaystyle 2^{1+13-16}=}$${\displaystyle 2^{-2}=}$${\displaystyle \frac{1}{4}}$;

9) ${\displaystyle 2\cdot 2^{-5}\cdot 2^6\cdot 2^0=}$${\displaystyle 2^{1-5+6+0}=}$${\displaystyle 2^2=}$${\displaystyle 4}$.