228 Вправа 228Алгебра 8 клас Бевз 2025 · § 5 · стор. 51

228. Виконай множення.

а) ${\displaystyle \frac{a-b}{3}\cdot \frac{15}{a^2-b^2}=}$${\displaystyle \frac{(a-b)\cdot 15}{3\cdot (a^2-b^2)}=}$${\displaystyle \frac{5\cdot \cancel{3}\cancel{\left(a-b\right)}}{\cancel{3}(a+b)\cancel{\left(a-b\right)}}=}$${\displaystyle \frac{5}{a+b}}$;

б) ${\displaystyle \frac{a-x}{4m^2}\cdot \frac{16am}{{(a-x)}^2}=}$${\displaystyle \frac{(a-x)\cdot 16am}{4m^2\cdot {(a-x)}^2}}$=

=${\displaystyle \frac{4\cdot \cancel{4}a\cancel{m}\cancel{\left(a-x\right)}}{\cancel{4}\cancel{m}m\cancel{\left(a-x\right)}(a-x)}=}$${\displaystyle \frac{4a}{m\left(a-x\right)}}$;

в) ${\displaystyle \frac{a^2-c^2}{2ac}\cdot \frac{8}{{(a+c)}^2}=}$${\displaystyle \frac{(a^2-c^2)\cdot 8}{2ac\cdot {(a+c)}^2}}$=

=${\displaystyle \frac{4\cdot \cancel{2}\cancel{\left(a+c\right)}(a-c)}{\cancel{2}ac\cancel{\left(a+c\right)}(a+c)}=}$${\displaystyle \frac{4\left(a-c\right)}{ac\left(a+c\right)}}$;

г) ${\displaystyle \frac{3a-3b}{4a+4c}\cdot \frac{{(a+c)}^2}{a^2-b^2}=}$${\displaystyle \frac{(3a-3b)\cdot {(a+c)}^2}{(4a+4c)\cdot (a^2-b^2)}}$=

=${\displaystyle \frac{3\cancel{\left(a+c\right)}(a+c)\cancel{\left(a-b\right)}}{4(a+b)\cancel{\left(a+c\right)}\cancel{\left(a-b\right)}}=}$${\displaystyle \frac{3\left(a+c\right)}{4\left(a+b\right)}}$;

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

=${\displaystyle \frac{6\cancel{\left(a+3\right)}\cancel{\left(a-2\right)}}{(a+2)\cancel{\left(a+3\right)}(a+3)\cancel{\left(a-2\right)}}=}$${\displaystyle \frac{6}{\left(a+2\right)\left(a+3\right)}}$;

д) ${\displaystyle \frac{x^2-y^2}{12}\cdot \frac{6}{{(x-y)}^2}=}$${\displaystyle \frac{(x^2-y^2)\cdot 6}{12\cdot {(x-y)}^2}}$=

=${\displaystyle \frac{\cancel{6}(x+y)\cancel{\left(x-y\right)}}{2\cdot \cancel{6}\cancel{\left(x-y\right)}(x-y)}=}$${\displaystyle \frac{x+y}{2\left(x-y\right)}}$.