297 Вправа 297Алгебра 8 клас Тарасенкова 2025 · § 8 · стор. 86

297. Знайдіть a⁻², якщо a дорівнює:

Мішане число й десятковий дріб спочатку записуємо звичайним дробом, потім «перевертаємо».

1) ${\displaystyle {\left(1\frac{2}{3}\right)}^{-2}=}$${\displaystyle {\left(\frac{5}{3}\right)}^{-2}=}$${\displaystyle {\left(\frac{3}{5}\right)}^2=}$${\displaystyle \frac{9}{25}}$;

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

3) ${\displaystyle {(-2\frac{1}{5})}^{-2}=}$${\displaystyle {(-\frac{11}{5})}^{-2}=}$${\displaystyle {\left(-\frac{5}{11}\right)}^2=}$${\displaystyle \frac{25}{121}}$;

4) ${\displaystyle {(-3\frac{1}{8})}^{-2}=}$${\displaystyle {(-\frac{25}{8})}^{-2}=}$${\displaystyle {\left(-\frac{8}{25}\right)}^2=}$${\displaystyle \frac{64}{625}}$;

5) ${\displaystyle {0,2}^{-2}=}$${\displaystyle {\left(\frac{1}{5}\right)}^{-2}=}$${\displaystyle 5^2=}$${\displaystyle 25}$;

6) ${\displaystyle {0,05}^{-2}=}$${\displaystyle {\left(\frac{1}{20}\right)}^{-2}=}$${\displaystyle {20}^2=}$${\displaystyle 400}$;

7) ${\displaystyle {1,2}^{-2}=}$${\displaystyle {\left(\frac{6}{5}\right)}^{-2}=}$${\displaystyle {\left(\frac{5}{6}\right)}^2=}$${\displaystyle \frac{25}{36}}$;

8) ${\displaystyle {(-1,2)}^{-2}=}$${\displaystyle {(-\frac{6}{5})}^{-2}=}$${\displaystyle {\left(-\frac{5}{6}\right)}^2=}$${\displaystyle \frac{25}{36}}$.