Упр.1281 ГДЗ Алимов 10-11 класс (Алгебра)

$$\frac{x^{\frac12}}{1+x^{\frac12}}\cdot\left(\frac{x^{\frac12}}{1-x^{\frac12}}-\frac{1}{x^{\frac12}-x}\right)$$

Преобразуем вторую дробь:
$$x^{\frac12}-x=x^{\frac12}(1-x^{\frac12})$$

Тогда
$$\frac{x^{\frac12}}{1+x^{\frac12}}\cdot\left(\frac{x^{\frac12}}{1-x^{\frac12}}-\frac{1}{x^{\frac12}(1-x^{\frac12})}\right) =\frac{x^{\frac12}}{1+x^{\frac12}}\cdot\frac{x-1}{x^{\frac12}(1-x^{\frac12})}$$

Сокращаем:
$$\frac{x-1}{(1+x^{\frac12})(1-x^{\frac12})}=\frac{x-1}{1-x}=-1$$

Ответ: $$-1$$

$$\frac{m+2m^{\frac12}+1}{2m^{\frac12}}\cdot\left(\frac{2m^{\frac12}}{m^{\frac12}-1}-\frac{4m^{\frac12}}{m-1}\right)$$

Заметим, что
$$m+2m^{\frac12}+1=\left(m^{\frac12}+1\right)^2,\qquad m-1=\left(m^{\frac12}-1\right)\left(m^{\frac12}+1\right).$$

Тогда
$$\frac{\left(m^{\frac12}+1\right)^2}{2m^{\frac12}}\cdot\left(\frac{2m^{\frac12}}{m^{\frac12}-1}-\frac{4m^{\frac12}}{(m^{\frac12}-1)(m^{\frac12}+1)}\right)$$

Приведём к общему знаменателю:
$$\frac{\left(m^{\frac12}+1\right)^2}{2m^{\frac12}}\cdot\frac{2m^{\frac12}(m^{\frac12}+1)-4m^{\frac12}}{(m^{\frac12}-1)(m^{\frac12}+1)}$$

$$=\frac{\left(m^{\frac12}+1\right)^2}{2m^{\frac12}}\cdot\frac{2m^{\frac12}-2m^{\frac12}+2}{m^{\frac12}-1} =\frac{\left(m^{\frac12}+1\right)^2}{2m^{\frac12}}\cdot\frac{2(m^{\frac12}-1)}{m^{\frac12}-1}$$

После сокращения получаем:
$$m^{\frac12}+1$$

Ответ: $$m^{\frac12}+1$$