457 Вправа 457Алгебра 8 клас Бевз 2025 · § 9 · стор. 92

457. Спрости вираз.

а) ${\displaystyle 0,2x^{-3}\cdot 5x^2{y}^3=}$${\displaystyle \left(0,2\cdot 5\right)\cdot x^{-3+2}\cdot y^3}$=

=${\displaystyle x^{-1}{y}^3=}$${\displaystyle \frac{y^3}{x}}$;

б) ${\displaystyle 3^{-3}{a}^{-1}x\cdot 81a^{2}x=}$${\displaystyle \frac{1}{27}{a}^{-1}x\cdot 81a^{2}x}$=

=${\displaystyle \left(\frac{1}{27}\cdot 81\right)\cdot a^{-1+2}\cdot x^{1+1}=}$${\displaystyle 3a{x}^2}$;

в) ${\displaystyle 0,2c^{-5}x:\left(5^{-2}{c}^{-3}{x}^{-1}\right)}$=

=${\displaystyle 0,2c^{-5}x:\frac{1}{25}{c}^{-3}{x}^{-1}}$=

=${\displaystyle \left(0,2\cdot 25\right)\cdot c^{-5+3}\cdot x^{1+1}=}$${\displaystyle 5c^{-2}{x}^2=}$${\displaystyle \frac{5x^2}{c^2}}$;

г) ${\displaystyle \frac{27x^{-1}{y}^2}{10}\cdot \frac{5x^6}{9y^{-4}}}$=

=${\displaystyle \frac{27\cdot 5}{10\cdot 9}\cdot x^{-1+6}\cdot y^{2+4}=}$${\displaystyle 1,5x^5{y}^6}$;

ґ) ${\displaystyle \frac{3x^8}{4y^3{z}^{-2}}:\frac{x^7{y}^{-4}}{12z^{-2}}}$=

=${\displaystyle \frac{3\cdot 12}{4}\cdot x^{8-7}\cdot y^{-3+4}\cdot z^{2-2}=}$${\displaystyle 9xy}$;

д) ${\displaystyle {\left(\frac{x^2{y}^{-3}}{6z}\right)}^{-3}:{\left(\frac{x^3{y}^{-5}}{9z}\right)}^{-2}=}$${\displaystyle \frac{{\left(6z\right)}^3}{{\left(x^2{y}^{-3}\right)}^3}:\frac{{\left(9z\right)}^2}{{\left(x^3{y}^{-5}\right)}^2}}$=

=${\displaystyle \frac{216z^3}{x^6{y}^{-9}}:\frac{81z^2}{x^6{y}^{-10}}=}$${\displaystyle \frac{216}{81}\cdot x^{-6+6}\cdot y^{9-10}\cdot z^{3-2}}$=

=${\displaystyle \frac{8}{3}{y}^{-1}z=}$${\displaystyle \frac{8z}{3y}}$;

е) ${\displaystyle {\left(2a^{-2}{x}^3\right)}^2\cdot {\left(\frac{a}{x}\right)}^6=}$${\displaystyle 4a^{-4}{x}^6\cdot \frac{a^6}{x^6}}$=

=${\displaystyle 4\cdot a^{-4+6}\cdot x^{6-6}=}$${\displaystyle 4a^2}$;

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

=${\displaystyle \frac{1}{2}\cdot x^{-1-2}\cdot y^{1+2}=}$${\displaystyle 0,5x^{-3}{y}^3=}$${\displaystyle \frac{y^3}{2x^3}}$.