208* Вправа 208Алгебра 8 клас Тарасенкова 2025 · § 5 · стор. 59

208. Спростіть вираз:

1) Дія в дужках (${\displaystyle xy-y^2=}$${\displaystyle y(x-y)}$;

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

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

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

${\displaystyle \frac{y}{x+y}+\frac{x}{x+y}=}$${\displaystyle \frac{x+y}{x+y}=}$${\displaystyle 1}$;

2) ${\displaystyle 4-2y-2x+xy=}$${\displaystyle 2(2-y)-x(2-y)=}$${\displaystyle (x-2)(y-2)}$;

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

дія в дужках:

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

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