15 Вправа 15 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 318

15. Спростіть вираз:

1) ${\displaystyle a^{-10}:a^{-6}\cdot a^{-5}=}$${\displaystyle a^{-10-(-6)+(-5)}=}$${\displaystyle a^{-10+6-5}=}$${\displaystyle a^{-9}}$;

2) ${\displaystyle a^{-10}:a^6\cdot a^{-5}=}$${\displaystyle a^{-10-6+(-5)}=}$${\displaystyle a^{-21}}$;

3) ${\displaystyle a^{-10}:a^{-6}:a^{-5}=}$${\displaystyle a^{-10-(-6)-(-5)}=}$${\displaystyle a^{-10+6+5}=}$${\displaystyle a^1=}$${\displaystyle a}$;

4) ${\displaystyle a^{-10}\cdot a^{-6}\cdot a^{-5}=}$${\displaystyle a^{-10-6-5}=}$${\displaystyle a^{-21}}$;

5) ${\displaystyle a^0:a^{-8}\cdot a^{-12}=}$${\displaystyle a^{0-(-8)+(-12)}=}$${\displaystyle a^{8-12}=}$${\displaystyle a^{-4}}$;

6) ${\displaystyle a^0:(a^{-8}\cdot a^{-12})=}$${\displaystyle a^0:a^{-20}=}$${\displaystyle a^{0-(-20)}=}$${\displaystyle a^{20}}$;

7) ${\displaystyle {\left(a^{-4}\right)}^{-3}:{\left(a^3\right)}^4=}$${\displaystyle a^{12}:a^{12}=}$${\displaystyle a^0=}$${\displaystyle 1}$;

8) ${\displaystyle {\left(a^{-4}\right)}^{-3}\cdot {\left(a^3\right)}^4=}$${\displaystyle a^{12}\cdot a^{12}=}$${\displaystyle a^{24}}$;

9) ${\displaystyle {\left(a^{-10}{a}^{40}\right)}^{-3}:{\left(a^{10}{a}^{-30}\right)}^0=}$${\displaystyle {\left(a^{30}\right)}^{-3}:1=}$${\displaystyle a^{-90}}$;

10) ${\displaystyle {\left(a^{-10}{a}^{40}\right)}^0:{\left(a^{10}{a}^{-30}\right)}^0=}$${\displaystyle 1:1=}$${\displaystyle 1}$.