303 Вправа 303Алгебра 8 клас Тарасенкова 2025 · § 8 · стор. 86

303. Обчисліть:

1) ${\displaystyle 5\cdot 9^0+{(1,5)}^{-1}=}$${\displaystyle 5\cdot 1+{\left(\frac{3}{2}\right)}^{-1}=}$${\displaystyle 5+\frac{2}{3}=}$${\displaystyle 5\frac{2}{3}}$;

2) ${\displaystyle {\left(5\frac{1}{3}\right)}^0\cdot {\left(1\frac{1}{3}\right)}^{-1}+2\cdot {\left(\frac{1}{5}\right)}^{-2}=}$${\displaystyle 1\cdot {\left(\frac{4}{3}\right)}^{-1}+2\cdot 5^2=}$${\displaystyle \frac{3}{4}+2\cdot 25=}$${\displaystyle 0,75+50=}$${\displaystyle 50,75}$;

3) ${\displaystyle {\left(\frac{1}{3}\right)}^{-3}:{\left(\frac{1}{6}\right)}^{-2}\cdot 2^{-4}=}$${\displaystyle 3^3:6^2\cdot \frac{1}{2^4}=}$${\displaystyle 27:36\cdot \frac{1}{16}=}$${\displaystyle \frac{27}{36}\cdot \frac{1}{16}=}$${\displaystyle \frac{3}{4}\cdot \frac{1}{16}=}$${\displaystyle \frac{3}{64}}$;

4) ${\displaystyle -{(-2,5)}^{-2}\cdot {\left(3\frac{1}{8}\right)}^{-1}=}$${\displaystyle -{(-\frac{5}{2})}^{-2}\cdot {\left(\frac{25}{8}\right)}^{-1}=}$${\displaystyle -{\left(-\frac{2}{5}\right)}^2\cdot \frac{8}{25}=}$${\displaystyle -\frac{4}{25}\cdot \frac{8}{25}=}$${\displaystyle -\frac{32}{625}}$.