Упр.9.30 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

Задача

1) (a^(1/6)+1)/(a^(1/3)-1);
2) (vm-(mn)^(1/4))/((mn)^(1/4)-vn);
3) (a-b)/(a^(1/3)-b^(1/3));
4) (avb-bva)/v(ab);
5) ((ab)^(1/3)+a^(1/3))/((a^2 b^2)^(1/3)+(a^2 b)^(1/3));
6) (3+3^(1/4))/3^(1/4).

Подробный ответ

$$\frac{\sqrt[6]{a}+1}{\sqrt[3]{a}-1}=\frac{\sqrt[6]{a}+1}{\left(\sqrt[6]{a}\right)^2-1}=\frac{\sqrt[6]{a}+1}{(\sqrt[6]{a}-1)(\sqrt[6]{a}+1)}=\frac{1}{\sqrt[6]{a}-1}.$$
$$\frac{\sqrt{m}-\sqrt[4]{mn}}{\sqrt[4]{mn}-\sqrt{n}}=\frac{\sqrt[4]{m^2}-\sqrt[4]{mn^2}}{\sqrt[4]{mn}-\sqrt[4]{n^2}}=\frac{\sqrt[4]{m}\,(\sqrt[4]{m}-\sqrt[4]{n})}{\sqrt[4]{n}\,(\sqrt[4]{m}-\sqrt[4]{n})}=\frac{\sqrt[4]{m}}{\sqrt[4]{n}}=\sqrt[4]{\frac{m}{n}}.$$
$$\frac{a-b}{\sqrt[3]{a}-\sqrt[3]{b}}=\frac{(\sqrt[3]{a})^3-(\sqrt[3]{b})^3}{\sqrt[3]{a}-\sqrt[3]{b}}=\frac{(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2})}{\sqrt[3]{a}-\sqrt[3]{b}}.$$
$$\frac{a-b}{\sqrt[3]{a}-\sqrt[3]{b}}=\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}.$$
$$\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}=\frac{\sqrt{a^2b}-\sqrt{ab^2}}{\sqrt{ab}}=\frac{\sqrt{ab}\,(\sqrt{a}-\sqrt{b})}{\sqrt{ab}}=\sqrt{a}-\sqrt{b}.$$
$$\frac{\sqrt[3]{ab}+\sqrt[3]{a}}{\sqrt[3]{a^2b^2}+\sqrt[3]{a^2b}}=\frac{\sqrt[3]{a}\,(\sqrt[3]{b}+1)}{\sqrt[3]{a^2b}\,(\sqrt[3]{b}+1)}=\frac{\sqrt[3]{a}}{\sqrt[3]{a^2b}}=\sqrt[3]{\frac{1}{ab}}=\frac{1}{\sqrt[3]{ab}}.$$
$$\frac{3+\sqrt[4]{3}}{\sqrt[4]{3}}=\frac{\sqrt[4]{3^4}+\sqrt[4]{3}}{\sqrt[4]{3}}=\frac{\sqrt[4]{3}\,(\sqrt[4]{3^3}+1)}{\sqrt[4]{3}}=\sqrt[4]{27}+1.$$