230 Вправа 230Алгебра 8 клас Бевз 2025 · § 5 · стор. 51

230. Виконай множення.

а) ${\displaystyle \frac{x^2-c^2}{x^2-a^2}\cdot \frac{a+x}{c+x}=}$${\displaystyle \frac{(x^2-c^2)\cdot (a+x)}{(x^2-a^2)\cdot (c+x)}}$=

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

б) ${\displaystyle \frac{{(a+b)}^2}{(a-b)b}\cdot \frac{b}{a^2-b^2}=}$${\displaystyle \frac{{(a+b)}^2\cdot b}{(a-b)b\cdot (a^2-b^2)}}$=

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

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

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