376° Вправа 376Алгебра 8 клас Тарасенкова 2025 · § 10 · стор. 106

376. Запишіть як степінь з натуральним показником:

${\displaystyle {\left(\frac{A}{B}\right)}^{-1}=}$${\displaystyle \frac{B}{A}}$ — дріб «перевертаємо».

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

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

3) ${\displaystyle {\left(\frac{1}{abc}\right)}^{-1}=}$${\displaystyle abc}$;

4) ${\displaystyle {\left(\frac{1}{m+n}\right)}^{-1}=}$${\displaystyle m+n}$;

5) ${\displaystyle {\left(\frac{1}{ac-p}\right)}^{-1}=}$${\displaystyle ac-p}$;

6) ${\displaystyle {\left(\frac{1}{pm-n^2}\right)}^{-1}=}$${\displaystyle pm-n^2}$;

7) ${\displaystyle {\left(\frac{a}{a+8}\right)}^{-1}=}$${\displaystyle \frac{a+8}{a}}$;

8) ${\displaystyle {\left(\frac{x}{zy}\right)}^{-1}=}$${\displaystyle \frac{zy}{x}}$;

9) ${\displaystyle {\left(\frac{cd}{ab}\right)}^{-1}=}$${\displaystyle \frac{ab}{cd}}$;

10) ${\displaystyle {\left(\frac{n-m}{m+n}\right)}^{-1}=}$${\displaystyle \frac{m+n}{n-m}}$;

11) ${\displaystyle {\left(\frac{x^2-1}{x^2+1}\right)}^{-1}=}$${\displaystyle \frac{x^2+1}{x^2-1}}$;

12) ${\displaystyle {\left(\frac{n^2+m^2}{m^2-n^2}\right)}^{-1}=}$${\displaystyle \frac{m^2-n^2}{n^2+m^2}}$.