298 Вправа 298Алгебра 8 клас Тарасенкова 2025 · § 8 · стор. 86

298. Обчисліть:

1) ${\displaystyle 2^{-6}=}$${\displaystyle \frac{1}{2^6}=}$${\displaystyle \frac{1}{64}}$;

2) ${\displaystyle {(-1,1)}^{-2}=}$${\displaystyle {(-\frac{11}{10})}^{-2}=}$${\displaystyle {\left(-\frac{10}{11}\right)}^2=}$${\displaystyle \frac{100}{121}}$;

3) ${\displaystyle {(-4)}^{-2}=}$${\displaystyle \frac{1}{{(-4)}^2}=}$${\displaystyle \frac{1}{16}}$;

4) ${\displaystyle {0,1}^{-2}=}$${\displaystyle {\left(\frac{1}{10}\right)}^{-2}=}$${\displaystyle {10}^2=}$${\displaystyle 100}$;

5) ${\displaystyle {0,5}^{-3}=}$${\displaystyle {\left(\frac{1}{2}\right)}^{-3}=}$${\displaystyle 2^3=}$${\displaystyle 8}$;

6) ${\displaystyle {(-2)}^{-5}=}$${\displaystyle \frac{1}{{(-2)}^5}=}$${\displaystyle \frac{1}{-32}=}$${\displaystyle -\frac{1}{32}}$;

7) ${\displaystyle {(-3)}^{-3}=}$${\displaystyle \frac{1}{{(-3)}^3}=}$${\displaystyle -\frac{1}{27}}$;

8) ${\displaystyle 3^{-3}=}$${\displaystyle \frac{1}{3^3}=}$${\displaystyle \frac{1}{27}}$;

9) ${\displaystyle {(-0,2)}^{-3}=}$${\displaystyle {(-\frac{1}{5})}^{-3}=}$${\displaystyle {\left(-5\right)}^3=}$${\displaystyle -125}$;

10) ${\displaystyle {(-1,5)}^{-3}=}$${\displaystyle {(-\frac{3}{2})}^{-3}=}$${\displaystyle {\left(-\frac{2}{3}\right)}^3=}$${\displaystyle -\frac{8}{27}}$;

11) ${\displaystyle {\left(1\frac{1}{4}\right)}^{-2}=}$${\displaystyle {\left(\frac{5}{4}\right)}^{-2}=}$${\displaystyle {\left(\frac{4}{5}\right)}^2=}$${\displaystyle \frac{16}{25}}$;

12) ${\displaystyle {(-\frac{1}{6})}^{-3}=}$${\displaystyle {\left(-6\right)}^3=}$${\displaystyle -216}$.