Упр.23.30 ГДЗ Мерзляк Поляков 9 класс (Алгебра)

1) (x-y)/(x^(3/4)+x^(1/2) y^(1/4))·(x^(1/2) y^(1/4)+x^(1/4) y^(1/2))/(x^(1/2)+y^(1/2))·(x^(1/4) y^(-1/4))/(x^(1/2)-2x^(1/4) y^(1/4)+y^(1/2);
2) (m^(4/3)-27m^(1/3) n)/(m^(2/3)+3(mn)^(1/3)+9n^(2/3)):(1-3(n/m)^(1/3))-(m^2)^(1/3).

1. $$\frac{x-y}{x^{3/4}+x^{1/2}y^{1/4}}\cdot\frac{x^{1/2}y^{1/4}+x^{1/4}y^{1/2}}{x^{1/2}+y^{1/2}}\cdot\frac{x^{1/4}y^{-1/4}}{x^{1/2}-2x^{1/4}y^{1/4}+y^{1/2}}$$

   Представим выражения в знаменателях в виде произведений:

   $$x^{3/4}+x^{1/2}y^{1/4}=x^{1/2}\left(x^{1/4}+y^{1/4}\right),$$
   $$x^{1/2}y^{1/4}+x^{1/4}y^{1/2}=x^{1/4}y^{1/4}\left(x^{1/4}+y^{1/4}\right),$$
   $$x^{1/2}-2x^{1/4}y^{1/4}+y^{1/2}=\left(x^{1/4}-y^{1/4}\right)^2,$$
   $$x-y=\left(x^{1/2}-y^{1/2}\right)\left(x^{1/2}+y^{1/2}\right)=\left(x^{1/4}-y^{1/4}\right)\left(x^{1/4}+y^{1/4}\right)\left(x^{1/2}+y^{1/2}\right).$$

   Тогда
   $$\frac{\left(x^{1/4}-y^{1/4}\right)\left(x^{1/4}+y^{1/4}\right)\left(x^{1/2}+y^{1/2}\right)}{x^{1/2}\left(x^{1/4}+y^{1/4}\right)}\cdot\frac{x^{1/4}y^{1/4}\left(x^{1/4}+y^{1/4}\right)}{x^{1/2}+y^{1/2}}\cdot\frac{x^{1/4}y^{-1/4}}{\left(x^{1/4}-y^{1/4}\right)^2}.$$

   Сокращаем одинаковые множители:
   $$\frac{x^{1/4}-y^{1/4}}{x^{1/2}}\cdot x^{1/4}y^{1/4}\cdot\frac{x^{1/4}y^{-1/4}}{\left(x^{1/4}-y^{1/4}\right)^2} =\frac{x^{1/4}+y^{1/4}}{x^{1/4}-y^{1/4}}.$$

2. $$\frac{m^{4/3}-27m^{1/3}n}{m^{2/3}+3(mn)^{1/3}+9n^{2/3}}:\left(1-3\left(\frac{n}{m}\right)^{1/3}\right)-\sqrt[3]{m^2}$$

   Вынесем общий множитель в числителе и преобразуем дробь:
   $$m^{4/3}-27m^{1/3}n=m^{1/3}(m-27n),$$
   $$m^{2/3}+3(mn)^{1/3}+9n^{2/3}=m^{2/3}+3m^{1/3}n^{1/3}+9n^{2/3}.$$

   Заметим, что
   $$1-3\left(\frac{n}{m}\right)^{1/3}=\frac{m^{1/3}-3n^{1/3}}{m^{1/3}}.$$

   Тогда
   $$\frac{m^{1/3}(m-27n)}{m^{2/3}+3m^{1/3}n^{1/3}+9n^{2/3}}:\frac{m^{1/3}-3n^{1/3}}{m^{1/3}}-\sqrt[3]{m^2}.$$

   Используем разложение:
   $$m-27n=\left(m^{1/3}-3n^{1/3}\right)\left(m^{2/3}+3m^{1/3}n^{1/3}+9n^{2/3}\right).$$

   Тогда первая дробь сокращается:
   $$\frac{m^{1/3}\left(m^{1/3}-3n^{1/3}\right)\left(m^{2/3}+3m^{1/3}n^{1/3}+9n^{2/3}\right)}{m^{2/3}+3m^{1/3}n^{1/3}+9n^{2/3}}:\frac{m^{1/3}-3n^{1/3}}{m^{1/3}}=m^{2/3}.$$

   Получаем:
   $$m^{2/3}-\sqrt[3]{m^2}=m^{2/3}-m^{2/3}=0.$$

Ответ

1) $$\frac{x^{1/4}+y^{1/4}}{x^{1/4}-y^{1/4}}$$ 2) $$0$$