Упр.114 ГДЗ Макарычев Миндюк Феоктистов 8 класс (Алгебра)

$$x\ne0,\quad x\ne1$$

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

$$y\ne1,\quad y\ne-1$$

$$\frac{4y}{y^2-2y+1}:\left(\frac{4}{y^2-2y+1}-\frac{2}{y-1}+\frac{2}{y+1}\right)$$
$$=\frac{4y}{(y-1)^2}:\left(\frac{4(y+1)-2(y-1)(y+1)+2(y-1)^2}{(y-1)^2(y+1)}\right)$$
$$=\frac{4y}{(y-1)^2}:\frac{8}{(y-1)^2(y+1)} =\frac{4y}{(y-1)^2}\cdot\frac{(y-1)^2(y+1)}{8} =\frac{y(y+1)}{2}.$$

Ответ: $$\frac{2x+1}{x^2(x-1)^2};\qquad \frac{y(y+1)}{2}.$$