Упр.26.13 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

Упростите выражение:

1. $$\frac{a^{\frac{2}{3}}-2a^{\frac{1}{3}}b^{\frac{1}{3}}+b^{\frac{2}{3}}}{a^{\frac{4}{3}}-a^{\frac{2}{3}}b^{\frac{2}{3}}}:\frac{a^{\frac{1}{3}}b^{\frac{1}{3}}-b^{\frac{2}{3}}}{ab^{\frac{1}{3}}+a^{\frac{2}{3}}b^{\frac{2}{3}}};$$
2. $$\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}b^{\frac{1}{4}}}\cdot\frac{a-b}{\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)\left(a^{\frac{1}{2}}-a^{\frac{1}{4}}b^{\frac{1}{4}}\right)}.$$

1) $$\frac{a^{2/3}-2a^{1/3}b^{1/3}+b^{2/3}}{a^{4/3}-a^{2/3}b^{2/3}}:\frac{a^{1/3}b^{1/3}-b^{2/3}}{ab^{1/3}+a^{2/3}b^{2/3}}$$

Представим выражения в виде множителей:

$$a^{2/3}-2a^{1/3}b^{1/3}+b^{2/3}=\left(a^{1/3}-b^{1/3}\right)^2,$$

$$a^{4/3}-a^{2/3}b^{2/3}=a^{2/3}\left(a^{2/3}-b^{2/3}\right)=a^{2/3}\left(a^{1/3}-b^{1/3}\right)\left(a^{1/3}+b^{1/3}\right),$$

$$a^{1/3}b^{1/3}-b^{2/3}=b^{1/3}\left(a^{1/3}-b^{1/3}\right),$$

$$ab^{1/3}+a^{2/3}b^{2/3}=a^{2/3}b^{1/3}\left(a^{1/3}+b^{1/3}\right).$$

Тогда

$$\frac{\left(a^{1/3}-b^{1/3}\right)^2}{a^{2/3}\left(a^{1/3}-b^{1/3}\right)\left(a^{1/3}+b^{1/3}\right)}:\frac{b^{1/3}\left(a^{1/3}-b^{1/3}\right)}{a^{2/3}b^{1/3}\left(a^{1/3}+b^{1/3}\right)}$$

$$=\frac{a^{1/3}-b^{1/3}}{a^{2/3}\left(a^{1/3}+b^{1/3}\right)}\cdot\frac{a^{2/3}b^{1/3}\left(a^{1/3}+b^{1/3}\right)}{b^{1/3}\left(a^{1/3}-b^{1/3}\right)}=1.$$

2) $$\frac{a-b}{a^{3/4}+a^{1/2}b^{1/4}}\cdot\frac{a-b}{\left(a^{1/2}+b^{1/2}\right)\left(a^{1/2}-a^{1/4}b^{1/4}\right)}$$

Используем разложение $$a-b=\left(a^{1/2}-b^{1/2}\right)\left(a^{1/2}+b^{1/2}\right)$$ и преобразуем второй множитель:

$$a^{3/4}+a^{1/2}b^{1/4}=a^{1/2}\left(a^{1/4}+b^{1/4}\right),$$

$$a^{1/2}-a^{1/4}b^{1/4}=a^{1/4}\left(a^{1/4}-b^{1/4}\right).$$

Тогда

$$\frac{\left(a^{1/2}-b^{1/2}\right)\left(a^{1/2}+b^{1/2}\right)}{a^{1/2}\left(a^{1/4}+b^{1/4}\right)}\cdot\frac{\left(a^{1/2}-b^{1/2}\right)\left(a^{1/2}+b^{1/2}\right)}{\left(a^{1/2}+b^{1/2}\right)a^{1/4}\left(a^{1/4}-b^{1/4}\right)}$$

$$=\frac{\left(a^{1/2}-b^{1/2}\right)^2\left(a^{1/2}+b^{1/2}\right)}{a^{3/4}\left(a^{1/4}+b^{1/4}\right)\left(a^{1/4}-b^{1/4}\right)}.$$

Так как $$a^{1/2}-b^{1/2}=\left(a^{1/4}-b^{1/4}\right)\left(a^{1/4}+b^{1/4}\right),$$ получаем

$$\frac{\left(a^{1/4}-b^{1/4}\right)^2\left(a^{1/4}+b^{1/4}\right)^2\left(a^{1/2}+b^{1/2}\right)}{a^{3/4}\left(a^{1/4}+b^{1/4}\right)\left(a^{1/4}-b^{1/4}\right)}$$

$$=\frac{\left(a^{1/4}-b^{1/4}\right)\left(a^{1/4}+b^{1/4}\right)\left(a^{1/2}+b^{1/2}\right)}{a^{3/4}}.$$

Но $$\left(a^{1/4}-b^{1/4}\right)\left(a^{1/4}+b^{1/4}\right)=a^{1/2}-b^{1/2},$$ значит

$$\frac{\left(a^{1/2}-b^{1/2}\right)\left(a^{1/2}+b^{1/2}\right)}{a^{3/4}}=\frac{a-b}{a^{3/4}}=a^{1/4}-ba^{-3/4}.$$

Ответ: 1) $$1$$; 2) $$a^{1/4}-ba^{-3/4}$$.