392° Вправа 392Алгебра 8 клас Тарасенкова 2025 · § 10 · стор. 109

392. Спростіть вираз:

1) ${\displaystyle \frac{x}{y}\cdot {\left(\frac{2x}{y}\right)}^{-1}=}$${\displaystyle \frac{x}{y}\cdot \frac{y}{2x}=}$${\displaystyle \frac{\cancel{x}\cdot \cancel{y}}{\cancel{y}\cdot 2\cancel{x}}=}$${\displaystyle \frac{1}{2}}$;

2) ${\displaystyle \frac{p}{n}\cdot {\left(\frac{m}{n}\right)}^{-2}=}$${\displaystyle \frac{p}{n}\cdot \frac{n^2}{m^2}=}$${\displaystyle \frac{p\cdot \cancel{n}\cdot n}{\cancel{n}\cdot m^2}=}$${\displaystyle \frac{pn}{m^2}}$;

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

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

5) ${\displaystyle {\left(\frac{8}{b+1}\right)}^{-1}\cdot {\left(\frac{b+1}{2}\right)}^{-5}=}$${\displaystyle \frac{b+1}{8}\cdot \frac{2^5}{{(b+1)}^5}=}$${\displaystyle \frac{\cancel{\left(b+1\right)}\cdot 4\cdot \stackrel{1}{\cancel{8}}}{\underset{1}{\cancel{8}}\cdot \cancel{\left(b+1\right)}{(b+1)}^4}=}$${\displaystyle \frac{4}{{\left(b+1\right)}^4}}$;

6) ${\displaystyle {\left(\frac{10}{y+x}\right)}^{-1}\cdot {\left(\frac{y+x}{2}\right)}^{-3}=}$${\displaystyle \frac{y+x}{10}\cdot \frac{2^3}{{(y+x)}^3}=}$${\displaystyle \frac{\cancel{\left(y+x\right)}\cdot 4\cdot \stackrel{1}{\cancel{2}}}{5\cdot \underset{1}{\cancel{2}}\cdot \cancel{\left(y+x\right)}{(y+x)}^2}=}$${\displaystyle \frac{4}{5{\left(x+y\right)}^2}}$.