241 Вправа 241Алгебра 8 клас Бевз 2025 · § 5 · стор. 52

241. Спрости вираз.

а) ${\displaystyle \frac{x^2-9y^2}{x^2+6xy+9y^2}\cdot \frac{x^2+3xy}{xy-3y^2}=}$${\displaystyle \frac{(x^2-9y^2)\cdot (x^2+3xy)}{(x^2+6xy+9y^2)\cdot (xy-3y^2)}}$=

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

б) ${\displaystyle \frac{x^2+6x+9}{2x-4}\cdot \frac{x^2-4}{3x+9}=}$${\displaystyle \frac{(x^2+6x+9)\cdot (x^2-4)}{(2x-4)\cdot (3x+9)}}$=

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

в) ${\displaystyle \frac{2x^2-6x}{x^2+2xy+y^2}\cdot \frac{x^2-y^2}{x^2-3x}=}$${\displaystyle \frac{(2x^2-6x)\cdot (x^2-y^2)}{(x^2+2xy+y^2)\cdot (x^2-3x)}}$=

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

г) ${\displaystyle \frac{x^2-10x+25}{6x+3y}\cdot \frac{2xy+y^2}{25-x^2}=}$${\displaystyle \frac{(x^2-10x+25)\cdot (2xy+y^2)}{(6x+3y)\cdot (25-x^2)}}$=

=${\displaystyle -\frac{y\cancel{\left(2x+y\right)}\cancel{\left(x-5\right)}(x-5)}{3\cancel{\left(2x+y\right)}(x+5)\cancel{\left(x-5\right)}}=}$${\displaystyle -\frac{y(x-5)}{3(x+5)}=}$${\displaystyle \frac{y\left(5-x\right)}{3\left(5+x\right)}}$.