Упр.102 ГДЗ Алимов 10-11 класс (Алгебра)
Сравнить числа: 1) $$\sqrt[7]{\left(\frac{1}{2}-\frac{1}{3}\right)^2}$$ и $$\sqrt[7]{\left(\frac{1}{3}-\frac{1}{4}\right)^2}$$; 2) $$\sqrt[5]{\left(1*\frac{1}{4}-1*\frac{1}{5}\right)^3}$$ и $$\sqrt[5]{\left(1*\frac{1}{6}-1*\frac{1}{7}\right)^3}$$.
1) $$\sqrt[7]{\left(\frac12-\frac13\right)^2}=\sqrt[7]{\left(\frac16\right)^2}=\left(\frac16\right)^{\frac27},$$
$$\sqrt[7]{\left(\frac13-\frac14\right)^2}=\sqrt[7]{\left(\frac1{12}\right)^2}=\left(\frac1{12}\right)^{\frac27}.$$
Так как $$\frac16>\frac1{12},$$ то
$$\sqrt[7]{\left(\frac12-\frac13\right)^2}>\sqrt[7]{\left(\frac13-\frac14\right)^2}.$$
2) $$\sqrt[5]{\left(\frac14-\frac15\right)^3}=\sqrt[5]{\left(\frac1{20}\right)^3}=\left(\frac1{20}\right)^{\frac35},$$
$$\sqrt[5]{\left(\frac16-\frac17\right)^3}=\sqrt[5]{\left(\frac1{42}\right)^3}=\left(\frac1{42}\right)^{\frac35}.$$
Так как $$\frac1{20}>\frac1{42},$$ то
$$\sqrt[5]{\left(\frac14-\frac15\right)^3}>\sqrt[5]{\left(\frac16-\frac17\right)^3}.$$
Ответ:
1) $$\sqrt[7]{\left(\frac12-\frac13\right)^2}>\sqrt[7]{\left(\frac13-\frac14\right)^2};$$
2) $$\sqrt[5]{\left(\frac14-\frac15\right)^3}>\sqrt[5]{\left(\frac16-\frac17\right)^3}.$$
