ВАРИАНТ 2. 1) \(1\frac{1}{10}+\frac{1}{4}+2\frac{7}{10}+\frac{3}{4}\); 2) \(\frac{9}{20}\cdot\frac{3}{3}+\frac{5}{14}\cdot\frac{7}{25}\); 3) \(2\frac{1}{27}\cdot3\frac{1}{5}\cdot7\frac{4}{11}\); 4) \(9\left(x-3\frac{1}{2}\cdot\frac{4}{7}\right)=0\); 5) \(\left(\frac{1}{3}-\frac{5}{16}\right):\left(\frac{1}{6}-\frac{1}{8}\right)\); 6) \(\left(3\frac{1}{8}+1\frac{3}{4}\right)\cdot x=2\frac{1}{6}\).

1)

Сгруппируем целые и дробные части:

\(1\frac{1}{10}+2\frac{7}{10}=3\frac{8}{10}=3\frac{4}{5}\).

\(\frac{1}{4}+\frac{3}{4}=1\).

Тогда \(3\frac{4}{5}+1=4\frac{4}{5}\).

2)

\(\frac{9}{20}\cdot\frac{3}{3}+\frac{5}{14}\cdot\frac{7}{25}=\frac{9}{20}+\frac{35}{350}=\frac{9}{20}+\frac{1}{10}=\frac{11}{20}\).

3)

Преобразуем смешанные числа:

\(2\frac{1}{27}=\frac{55}{27}\), \(3\frac{1}{5}=\frac{16}{5}\), \(7\frac{4}{11}=\frac{81}{11}\).

\(\frac{55}{27}\cdot\frac{16}{5}\cdot\frac{81}{11}=48\).

4)

\(3\frac{1}{2}\cdot\frac{4}{7}=\frac{7}{2}\cdot\frac{4}{7}=2\).

Тогда \(9(x-2)=0\), поэтому \(x-2=0\text{ и }x=2\).

5)

\(\frac{1}{3}-\frac{5}{16}=\frac{16-15}{48}=\frac{1}{48}\).

\(\frac{1}{6}-\frac{1}{8}=\frac{4-3}{24}=\frac{1}{24}\).

\(\frac{1}{48}:\frac{1}{24}=\frac{1}{48}\cdot24=\frac{1}{2}\).

6)

\(3\frac{1}{8}+1\frac{3}{4}=\frac{25}{8}+\frac{14}{8}=\frac{39}{8}\).

Получаем \(\frac{39}{8}x=2\frac{1}{6}=\frac{13}{6}\).

\(x=\frac{13}{6}\cdot\frac{8}{39}=\frac{4}{9}\).

Ответ: 1) \(4\frac{4}{5}\); 2) \(\frac{11}{20}\); 3) \(48\); 4) \(x=2\); 5) \(\frac{1}{2}\); 6) \(x=\frac{4}{9}\).