279 Вправа 279Алгебра 8 клас Бевз 2025 · § 6 · стор. 59

279. Виконай ділення.

а) ${\displaystyle \frac{16x{y}^2-x^3}{x^2-2xy+y^2}:\frac{x^{2}y-16y^3}{x^2-y^2}=}$${\displaystyle \frac{16x{y}^2-x^3}{x^2-2xy+y^2}\cdot \frac{x^2-y^2}{x^{2}y-16y^3}}$=

=${\displaystyle \frac{(16x{y}^2-x^3)\cdot (x^2-y^2)}{(x^2-2xy+y^2)\cdot (x^{2}y-16y^3)}}$=

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

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

б) ${\displaystyle \frac{x^2+2xy+y^2}{y^{2}x-9x^3}:\frac{x^2-y^2}{9x^{2}y-y^3}=}$${\displaystyle \frac{x^2+2xy+y^2}{y^{2}x-9x^3}\cdot \frac{9x^{2}y-y^3}{x^2-y^2}}$=

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

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

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

в) ${\displaystyle \frac{x^2+xy}{5x^2-5y^2}:\frac{x^2-xy}{x^3+y^3}=}$${\displaystyle \frac{x^2+xy}{5x^2-5y^2}\cdot \frac{x^3+y^3}{x^2-xy}}$=

=${\displaystyle \frac{(x^2+xy)\cdot (x^3+y^3)}{(5x^2-5y^2)\cdot (x^2-xy)}}$=

=${\displaystyle \frac{\cancel{x}\cancel{\left(x+y\right)}(x+y)(x^2-xy+y^2)}{5\cancel{x}\cancel{\left(x+y\right)}{(x-y)}^2}=}$${\displaystyle \frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{5{\left(x-y\right)}^2}}$;

г) ${\displaystyle \frac{x^2-2xy}{x^2+4y^2}:\frac{{(x-2y)}^2}{x^4-16y^4}=}$${\displaystyle \frac{x^2-2xy}{x^2+4y^2}\cdot \frac{x^4-16y^4}{{(x-2y)}^2}}$=

=${\displaystyle \frac{(x^2-2xy)\cdot (x^4-16y^4)}{(x^2+4y^2)\cdot {(x-2y)}^2}}$=

=${\displaystyle \frac{x(x+2y)\cancel{\left(x^2+4y^2\right)}\cancel{{\left(x-2y\right)}^2}}{\cancel{\left(x^2+4y^2\right)}\cancel{{\left(x-2y\right)}^2}}=}$${\displaystyle x\left(x+2y\right)}$.