269 Вправа 269Алгебра 8 клас Бевз 2025 · § 6 · стор. 58

269. Знайди частку.

а) ${\displaystyle \frac{a^3+a^2}{11c^2}:\frac{4a+4}{c^3}=}$${\displaystyle \frac{a^3+a^2}{11c^2}\cdot \frac{c^3}{4a+4}=}$${\displaystyle \frac{(a^3+a^2)\cdot c^3}{11c^2\cdot (4a+4)}}$=

=${\displaystyle \frac{a^2\cancel{c^2}c\cancel{\left(a+1\right)}}{44\cancel{c^2}\cancel{\left(a+1\right)}}=}$${\displaystyle \frac{a^{2}c}{44}}$;

б) ${\displaystyle \frac{8cx}{c^2-2c}:\frac{4cx}{3c-6}=}$${\displaystyle \frac{8cx}{c^2-2c}\cdot \frac{3c-6}{4cx}=}$${\displaystyle \frac{8cx\cdot (3c-6)}{(c^2-2c)\cdot 4cx}}$=

=${\displaystyle \frac{6\cdot \cancel{4}\cancel{c}\cancel{x}\cancel{\left(c-2\right)}}{\cancel{4}\cancel{c}c\cancel{x}\cancel{\left(c-2\right)}}=}$${\displaystyle \frac{6}{c}}$;

в) ${\displaystyle \frac{m{c}^2}{m^2-1}:\frac{3c}{m^3-m^2}=}$${\displaystyle \frac{m{c}^2}{m^2-1}\cdot \frac{m^3-m^2}{3c}=}$${\displaystyle \frac{m{c}^2\cdot (m^3-m^2)}{(m^2-1)\cdot 3c}}$=

=${\displaystyle \frac{m^3\cancel{c}c\cancel{\left(m-1\right)}}{3\cancel{c}(m+1)\cancel{\left(m-1\right)}}=}$${\displaystyle \frac{m^{3}c}{3\left(m+1\right)}}$.