391° Вправа 391Алгебра 8 клас Тарасенкова 2025 · § 10 · стор. 109

391. Спростіть вираз:

1) ${\displaystyle {(a+b)}^{-5}\cdot {(a-b)}^{-5}=}$${\displaystyle {\left((a+b)(a-b)\right)}^{-5}=}$${\displaystyle {\left(a^2-b^2\right)}^{-5}}$;

2) ${\displaystyle {\left(\frac{2}{z}\right)}^{-2}\cdot {\left(\frac{z}{3}\right)}^{-2}=}$${\displaystyle {\left(\frac{2\cdot \cancel{z}}{\cancel{z}\cdot 3}\right)}^{-2}=}$${\displaystyle {\left(\frac{2}{3}\right)}^{-2}=}$${\displaystyle {\left(\frac{3}{2}\right)}^2=}$${\displaystyle \frac{9}{4}=}$${\displaystyle 2,25}$;

3) ${\displaystyle {\left(\frac{n}{n+1}\right)}^{-5}\cdot {\left(\frac{n+1}{n}\right)}^{-5}=}$${\displaystyle {\left(\frac{\cancel{n}\cdot \cancel{\left(n+1\right)}}{\cancel{\left(n+1\right)}\cdot \cancel{n}}\right)}^{-5}=}$${\displaystyle 1^{-5}=}$${\displaystyle 1}$;

4) ${\displaystyle {\left(\frac{b}{c+a}\right)}^{-3}\cdot {\left(\frac{2a+2c}{b}\right)}^{-3}=}$${\displaystyle {\left(\frac{\cancel{b}\cdot 2\cancel{\left(a+c\right)}}{\cancel{\left(c+a\right)}\cdot \cancel{b}}\right)}^{-3}=}$${\displaystyle 2^{-3}=}$${\displaystyle \frac{1}{8}}$;

5) ${\displaystyle p^{-4}\cdot {\left(\frac{p+2}{p}\right)}^{-4}\cdot {\left(\frac{1}{p+2}\right)}^{-4}=}$${\displaystyle {\left(\frac{\cancel{p}\cdot \cancel{\left(p+2\right)}\cdot 1}{\cancel{p}\cdot \cancel{\left(p+2\right)}}\right)}^{-4}=}$${\displaystyle 1^{-4}=}$${\displaystyle 1}$;

6) ${\displaystyle {\left(\frac{x}{x+1}\right)}^{-1}:{\left(\frac{x^2}{x+1}\right)}^{-1}=}$${\displaystyle {(\frac{x}{x+1}:\frac{x^2}{x+1})}^{-1}=}$${\displaystyle {\left(\frac{\cancel{x}\cdot \cancel{\left(x+1\right)}}{\cancel{\left(x+1\right)}\cdot x\cdot \cancel{x}}\right)}^{-1}=}$${\displaystyle {\left(\frac{1}{x}\right)}^{-1}=}$${\displaystyle x}$.