Задание 142 Вариант 3 Самостоятельная работа ГДЗ Дидактические материалы Чесноков 6 класс (Математика)

6 chesnokov_didakt6 sam/var3/142 611

а) $$\frac{7}{12}y=1\frac{1}{4}$$

$$y=1\frac{1}{4}:\frac{7}{12} \\\\ y=\frac{5}{4}\cdot\frac{12}{7} \\\\ y=\frac{5\cdot 12}{4\cdot 7} \\\\ y=\frac{15}{7}=2\frac{1}{7}$$

Ответ: $$y=2\frac{1}{7}$$

б) $$3\frac{1}{15}-1\frac{14}{15}a=\frac{1}{6}$$

$$1\frac{14}{15}a=3\frac{1}{15}-\frac{1}{6} \\\\ 1\frac{14}{15}a=3\frac{2}{30}-\frac{5}{30} \\\\ 1\frac{14}{15}a=2\frac{27}{30}=2\frac{9}{10} \\\\ \frac{29}{15}a=2\frac{9}{10} \\\\ a=2\frac{9}{10}:\frac{29}{15} \\\\ a=\frac{29}{10}\cdot\frac{15}{29} \\\\ a=\frac{15}{10}=1{,}5$$

Ответ: $$a=1{,}5$$

в) $$\left(\frac{7}{18}+\frac{5}{24}z\right):3\frac{2}{3}=\frac{1}{3}$$

$$\frac{7}{18}+\frac{5}{24}z=\frac{1}{3}\cdot 3\frac{2}{3} \\\\ \frac{7}{18}+\frac{5}{24}z=\frac{1}{3}\cdot\frac{11}{3} \\\\ \frac{7}{18}+\frac{5}{24}z=\frac{11}{9} \\\\ \frac{5}{24}z=\frac{11}{9}-\frac{7}{18} \\\\ \frac{5}{24}z=\frac{22}{18}-\frac{7}{18} \\\\ \frac{5}{24}z=\frac{15}{18}=\frac{5}{6} \\\\ z=\frac{5}{6}:\frac{5}{24} \\\\ z=\frac{5}{6}\cdot\frac{24}{5}=4$$

Ответ: $$z=4$$

г) $$\frac{4}{7}x+\frac{5}{14}x-\frac{10}{21}x=\frac{1}{7}$$

$$\left(\frac{4}{7}+\frac{5}{14}-\frac{10}{21}\right)x=\frac{1}{7} \\\\ \left(\frac{24}{42}+\frac{15}{42}-\frac{20}{42}\right)x=\frac{1}{7} \\\\ \frac{19}{42}x=\frac{1}{7} \\\\ x=\frac{1}{7}:\frac{19}{42} \\\\ x=\frac{1}{7}\cdot\frac{42}{19}=\frac{6}{19}$$

Ответ: $$x=\frac{6}{19}$$

д) $$m-\frac{5}{12}m=\frac{1}{4}$$

$$\frac{7}{12}m=\frac{1}{4} \\\\ m=\frac{1}{4}:\frac{7}{12} \\\\ m=\frac{1}{4}\cdot\frac{12}{7}=\frac{3}{7}$$

Ответ: $$m=\frac{3}{7}$$