Решебник по математике 6 класс Мерзляк ФГОС Задание 421

 

$$\boxed{{\displaystyle \mathbf{421}\left(\mathbf{421}\right)\mathbf{.}Еуроки-ДЗбезмороки}}$$

Пояснение.

Решение.

$$a\ne 0;b\ne 0$$

$$1)\frac{3}{4}a=\frac{2}{3}\text{b }$$

$$a=\frac{2}{3}b:\frac{3}{4}$$

$$a=\frac{2}{3}\cdot \frac{4}{3}\cdot b$$

$$a=\frac{2\cdot 4}{3\cdot 3}\text{b }$$

$$a=\frac{8}{9}b⟹a<b.$$

$$2)\frac{2}{5}a=\frac{5}{7}\text{b }$$

$$a=\frac{5}{7}b:\frac{2}{5}$$

$$a=\frac{5}{7}\cdot \frac{5}{2}\cdot b$$

$$a=\frac{5\cdot 5}{7\cdot 2}\text{b }$$

$$a=\frac{25}{14}b⟹a>b.$$

$$\boxed{{\displaystyle \mathbf{421}.}}$$

$$\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\dots +\frac{1}{18}>\frac{1}{2}$$

$$\frac{1}{2}=\frac{9}{18}$$

$$\frac{9}{18}=\frac{1}{18}+\frac{1}{18}+\frac{1}{18}+\dots +\frac{1}{18}$$

$$\frac{1}{10}>\frac{1}{18};\frac{1}{11}>\frac{1}{18};\frac{1}{12}>\frac{1}{18};$$

$$\frac{1}{13}>\frac{1}{18};\frac{1}{14}>\frac{1}{18};\frac{1}{15}>\frac{1}{18};$$

$$\frac{1}{16}>\frac{1}{18};\frac{1}{17}>\frac{1}{18};\frac{1}{18}=\frac{1}{18}.$$

$$\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\dots +\frac{1}{18}>\frac{9}{18}$$

$$\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\dots +\frac{1}{18}>\frac{1}{2}$$

$$Чтоитребовалосьдоказать.$$