402 Вправа 402Алгебра 8 клас Тарасенкова 2025 · § 10 · стор. 111

402. Спростіть вираз:

1) ${\displaystyle \frac{27a^{-7}}{c^5}\cdot {\left(\frac{9a^{-10}}{c^3}\right)}^{-1}=}$${\displaystyle \frac{27a^{-7}}{c^5}\cdot \frac{c^3}{9a^{-10}}=}$${\displaystyle 3a^{-7+10}{c}^{3-5}=}$${\displaystyle 3a^3{c}^{-2}}$;

2) ${\displaystyle {\left(\frac{a^3}{2c^4{b}^{-8}}\right)}^4:{\left(\frac{a^{-6}}{0,5c^{-15}}\right)}^{-2}=}$${\displaystyle \frac{a^{12}{b}^{32}}{16c^{16}}:0,25a^{12}{c}^{-30}=}$${\displaystyle \frac{1}{16\cdot 0,25}{a}^{12-12}{b}^{32}{c}^{-16+30}=}$${\displaystyle 0,25b^{32}{c}^{14}}$;

3) ${\displaystyle {\left(\frac{a^7}{2m^{-1}{n}^{-2}}\right)}^{-6}:{\left(\frac{0,5n^5}{m^{-3}p}\right)}^{-2}=}$${\displaystyle {(0,5a^{7}m{n}^2)}^{-6}:{(0,5m^3{n}^5{p}^{-1})}^{-2}=}$${\displaystyle 64a^{-42}{m}^{-6}{n}^{-12}:4m^{-6}{n}^{-10}{p}^2=}$${\displaystyle 16a^{-42}{n}^{-2}{p}^{-2}}$;

4) ${\displaystyle {\left(\frac{4x^{-7}}{7y^{-9}{z}^2}\right)}^{-10}\cdot {\left(\frac{3\frac{1}{16}x}{y^{-1}z}\right)}^{-5}=}$${\displaystyle {\left(\frac{4}{7}\right)}^{-10}{x}^{70}{y}^{-90}{z}^{20}\cdot {\left(\frac{49}{16}\right)}^{-5}{x}^{-5}{y}^{-5}{z}^5=}$${\displaystyle \frac{7^{10}}{4^{10}}\cdot \frac{4^{10}}{7^{10}}\cdot x^{65}{y}^{-95}{z}^{25}=}$${\displaystyle x^{65}{y}^{-95}{z}^{25}}$.