Упр.78 ГДЗ Алимов 10-11 класс (Алгебра)

1)

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

Ответ: $$a$$

2)

$$\frac{b^{\frac{1}{5}}\left(\sqrt[5]{b^4}-\sqrt[5]{b^{-1}}\right)}{b^{\frac{2}{3}}\left(\sqrt[3]{b}-\sqrt[3]{b^{-2}}\right)} = \frac{b^{\frac{1}{5}}\left(b^{\frac{4}{5}}-b^{-\frac{1}{5}}\right)}{b^{\frac{2}{3}}\left(b^{\frac{1}{3}}-b^{-\frac{2}{3}}\right)} = \frac{b^{\frac{1}{5}}b^{-\frac{1}{5}}(b-1)}{b^{\frac{2}{3}}b^{-\frac{2}{3}}(b-1)} =1$$

Ответ: $$1$$

3)

$$\frac{a^{\frac{5}{3}}b^{-1}-a^{-\frac{1}{3}}}{\sqrt[3]{a^2}} = \frac{a^{-\frac{1}{3}}(a^2-b)}{a^{\frac{2}{3}}} = \frac{a^2-b}{a} \cdot \frac{1}{a^{\frac{1}{3}}} = \frac{a^2-b}{a^{\frac{4}{3}}}$$

Так как $$a^{\frac{4}{3}}=a\sqrt[3]{a},$$ получаем
$$\frac{a^{\frac{5}{3}}b^{-1}-a^{-\frac{1}{3}}}{\sqrt[3]{a^2}} = \frac{a^2-b}{ab}$$

Ответ: $$\frac{a^2-b}{ab}$$

4)

$$\frac{a^{\frac{1}{3}}\sqrt{b}+b^{\frac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}} = \frac{a^{\frac{1}{3}}b^{\frac{1}{2}}+b^{\frac{1}{3}}a^{\frac{1}{2}}}{a^{\frac{1}{6}}+b^{\frac{1}{6}}} = \frac{a^{\frac{2}{6}}b^{\frac{3}{6}}+b^{\frac{2}{6}}a^{\frac{3}{6}}}{a^{\frac{1}{6}}+b^{\frac{1}{6}}} = \frac{a^{\frac{2}{6}}b^{\frac{2}{6}}\left(a^{\frac{1}{6}}+b^{\frac{1}{6}}\right)}{a^{\frac{1}{6}}+b^{\frac{1}{6}}} = a^{\frac{2}{6}}b^{\frac{2}{6}}$$

$$a^{\frac{2}{6}}b^{\frac{2}{6}}=a^{\frac{1}{3}}b^{\frac{1}{3}}$$

Ответ: $$a^{\frac{1}{3}}b^{\frac{1}{3}}$$