229 Вправа 229Алгебра 8 клас Бевз 2025 · § 5 · стор. 51

229. Виконай множення.

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

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

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

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

в) ${\displaystyle \frac{x^2-1}{x^3-1}\cdot \frac{x^2+x+1}{x+1}=}$${\displaystyle \frac{(x^2-1)\cdot (x^2+x+1)}{(x^3-1)\cdot (x+1)}}$=

=${\displaystyle \frac{\cancel{\left(x+1\right)}\cancel{\left(x^2+x+1\right)}\cancel{\left(x-1\right)}}{\cancel{\left(x+1\right)}\cancel{\left(x^2+x+1\right)}\cancel{\left(x-1\right)}}=}$${\displaystyle 1}$.