450 Вправа 450Алгебра 8 клас Бевз 2025 · § 9 · стор. 91

450. Спрости вираз.

а) ${\displaystyle \frac{1}{6}{p}^2{q}^{-5}\cdot \frac{1}{2}{p}^{-1}{q}^{-3}=}$${\displaystyle \left(\frac{1}{6}\cdot \frac{1}{2}\right)\cdot p^{2-1}\cdot q^{-5-3}}$=

=${\displaystyle \frac{1}{12}p{q}^{-8}=}$${\displaystyle \frac{p}{12q^8}}$;

б) ${\displaystyle 15a{c}^{-2}:a^{2}c=}$${\displaystyle 15\cdot a^{1-2}\cdot c^{-2-1}}$=

=${\displaystyle 15a^{-1}{c}^{-3}=}$${\displaystyle \frac{15}{a{c}^3}}$;

в) ${\displaystyle 3,6x^4{y}^5:x{y}^5=}$${\displaystyle 3,6\cdot x^{4-1}\cdot y^{5-5}=}$${\displaystyle 3,6x^3}$;

г) ${\displaystyle \frac{6x^{-5}}{y^{-6}}\cdot \frac{y}{36x^{-7}}=}$${\displaystyle \frac{6}{36}\cdot x^{-5+7}\cdot y^{6+1}}$=

=${\displaystyle \frac{1}{6}{x}^2{y}^7=}$${\displaystyle \frac{x^2{y}^7}{6}}$;

ґ) ${\displaystyle \frac{5x^{-1}c}{3}\cdot \frac{9x^5}{c^{-3}}=}$${\displaystyle \frac{5\cdot 9}{3}\cdot c^{1+3}\cdot x^{-1+5}=}$${\displaystyle 15c^4{x}^4}$;

д) ${\displaystyle {\left(\frac{a^{-2}}{10^{2}x}\right)}^2\cdot {\left(a{x}^{-1}\right)}^2=}$${\displaystyle \frac{a^{-2\cdot 2}}{10^{2\cdot 2}{x}^2}\cdot a^2{x}^{-1\cdot 2}}$=

=${\displaystyle \frac{a^{-4}}{10^4{x}^2}\cdot a^2{x}^{-2}=}$${\displaystyle 10^{-4}\cdot a^{-4+2}\cdot x^{-2-2}}$=

=${\displaystyle 10^{-4}{a}^{-2}{x}^{-4}=}$${\displaystyle \frac{1}{10000a^2{x}^4}}$.