242 Вправа 242Алгебра 8 клас Бевз 2025 · § 5 · стор. 52

242. Спрости вираз.

а) ${\displaystyle \frac{xy-2y^2}{x^2+2xy}\cdot \frac{x^2+4xy+4y^2}{x^2-4y^2}=}$${\displaystyle \frac{(xy-2y^2)\cdot (x^2+4xy+4y^2)}{(x^2+2xy)\cdot (x^2-4y^2)}}$=

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

б) ${\displaystyle \frac{x^2-12x+36}{2x-8}\cdot \frac{2x^2-32}{x^2-6x}=}$${\displaystyle \frac{(x^2-12x+36)\cdot (2x^2-32)}{(2x-8)\cdot (x^2-6x)}}$=

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

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

=${\displaystyle \frac{\cancel{3}\cancel{x}y\cancel{\left(x+1\right)}(x+y)\cancel{\left(x-y\right)}}{\cancel{3}\cancel{x}\cancel{\left(x+1\right)}\cancel{\left(x-y\right)}(x-y)}=}$${\displaystyle \frac{y\left(x+y\right)}{x-y}}$;

г) ${\displaystyle \frac{x^3-81x}{x^2+14x+49}\cdot \frac{2x^2+14x}{x^3-9x^2}=}$${\displaystyle \frac{(x^3-81x)\cdot (2x^2+14x)}{(x^2+14x+49)\cdot (x^3-9x^2)}}$=

=${\displaystyle \frac{2\cancel{x^2}\cancel{\left(x+7\right)}(x+9)\cancel{\left(x-9\right)}}{\cancel{x^2}\cancel{\left(x+7\right)}(x+7)\cancel{\left(x-9\right)}}=}$${\displaystyle \frac{2\left(x+9\right)}{x+7}}$.