278 Вправа 278Алгебра 8 клас Бевз 2025 · § 6 · стор. 59

278. Виконай ділення.

а) ${\displaystyle \frac{x^2-xy}{x^2+xy}:\frac{xy}{x^{2}y+x{y}^2}=}$${\displaystyle \frac{x^2-xy}{x^2+xy}\cdot \frac{x^{2}y+x{y}^2}{xy}}$=

=${\displaystyle \frac{(x^2-xy)\cdot (x^{2}y+x{y}^2)}{(x^2+xy)\cdot xy}=}$${\displaystyle \frac{\cancel{x^2}\cancel{y}\cancel{\left(x+y\right)}(x-y)}{\cancel{x^2}\cancel{y}\cancel{\left(x+y\right)}}=}$${\displaystyle x-y}$;

б) ${\displaystyle \frac{a^{2}b-4b^3}{3a{b}^2}:\frac{a^2-2ab}{a^{2}b}=}$${\displaystyle \frac{a^{2}b-4b^3}{3a{b}^2}\cdot \frac{a^{2}b}{a^2-2ab}}$=

=${\displaystyle \frac{(a^{2}b-4b^3)\cdot a^{2}b}{3a{b}^2\cdot (a^2-2ab)}=}$${\displaystyle \frac{\cancel{a^2}\cancel{b^2}(a+2b)\cancel{\left(a-2b\right)}}{3\cancel{a^2}\cancel{b^2}\cancel{\left(a-2b\right)}}=}$${\displaystyle \frac{a+2b}{3}}$;

в) ${\displaystyle \frac{4x^2-16x+16}{x^2-9}:\frac{x-2}{x-3}=}$${\displaystyle \frac{4x^2-16x+16}{x^2-9}\cdot \frac{x-3}{x-2}}$=

=${\displaystyle \frac{(4x^2-16x+16)\cdot (x-3)}{(x^2-9)\cdot (x-2)}}$=

=${\displaystyle \frac{4\cancel{\left(x-2\right)}(x-2)\cancel{\left(x-3\right)}}{(x+3)\cancel{\left(x-2\right)}\cancel{\left(x-3\right)}}=}$${\displaystyle \frac{4\left(x-2\right)}{x+3}}$;

г) ${\displaystyle \frac{a^2+6a+9}{36-a^2}:\frac{a+3}{a+6}=}$${\displaystyle \frac{a^2+6a+9}{36-a^2}\cdot \frac{a+6}{a+3}}$=

=${\displaystyle \frac{(a^2+6a+9)\cdot (a+6)}{(36-a^2)\cdot (a+3)}=}$${\displaystyle -\frac{\cancel{\left(a+3\right)}(a+3)\cancel{\left(a+6\right)}}{\cancel{\left(a+3\right)}\cancel{\left(a+6\right)}(a-6)}}$=

=${\displaystyle -\frac{a+3}{a-6}=}$${\displaystyle \frac{a+3}{6-a}}$.