449 Вправа 449Алгебра 8 клас Бевз 2025 · § 9 · стор. 91

449. Спрости вираз.

а) ${\displaystyle 6x^{-2}c\cdot 1,5x{c}^{-3}=}$${\displaystyle \left(6\cdot 1,5\right)\cdot c^{1-3}\cdot x^{-2+1}}$=

=${\displaystyle 9c^{-2}{x}^{-1}=}$${\displaystyle \frac{9}{c^{2}x}}$;

б) ${\displaystyle 1,6x^{-1}{y}^{-5}\cdot \frac{5}{8}xy}$=

=${\displaystyle \left(1,6\cdot \frac{5}{8}\right)\cdot x^{-1+1}\cdot y^{-5+1}=}$${\displaystyle y^{-4}=}$${\displaystyle \frac{1}{y^4}}$;

в) ${\displaystyle \frac{3}{4}{a}^2{n}^{-4}\cdot 8a^{-3}{n}^2=}$${\displaystyle \left(\frac{3}{4}\cdot 8\right)\cdot a^{2-3}\cdot n^{-4+2}}$=

=${\displaystyle 6a^{-1}{n}^{-2}=}$${\displaystyle \frac{6}{a{n}^2}}$;

г) ${\displaystyle \frac{6x^{-5}}{a^{-6}}\cdot \frac{a}{36x^{-9}}=}$${\displaystyle \frac{6}{36}\cdot a^{6+1}\cdot x^{-5+9}}$=

=${\displaystyle \frac{1}{6}{a}^7{x}^4=}$${\displaystyle \frac{a^7{x}^4}{6}}$;

ґ) ${\displaystyle \frac{8x^2}{z}\cdot \frac{z^{-3}}{16x^{-3}}=}$${\displaystyle \frac{8}{16}\cdot x^{2+3}\cdot z^{-1-3}}$=

=${\displaystyle 0,5x^5{z}^{-4}=}$${\displaystyle \frac{x^5}{2z^4}}$;

д) ${\displaystyle \frac{14c^{-17}}{x^{-8}}\cdot \frac{x}{7c^{-18}}=}$${\displaystyle \frac{14}{7}\cdot c^{-17+18}\cdot x^{8+1}=}$${\displaystyle 2c{x}^9}$.