Решите уравнение: 1) \(2\frac{2}{11}x-\frac{5}{16}=1\frac{3}{4}\); 2) \(4\frac{2}{9}x+3\frac{5}{14}=6\frac{11}{21}\); 4) \(\frac{1}{3}x+\frac{1}{4}x+\frac{1}{5}x=1\frac{19}{75}\).

1) \(2\frac{2}{11}=\frac{24}{11}\), а \(1\frac{3}{4}=\frac{7}{4}\).

\[\frac{24}{11}x-\frac{5}{16}=\frac{7}{4}\]

Перенесём дробь \(\frac{5}{16}\) вправо:

\[\frac{24}{11}x=\frac{7}{4}+\frac{5}{16}=\frac{28}{16}+\frac{5}{16}=\frac{33}{16}\]

Разделим обе части на \(\frac{24}{11}\):

\[x=\frac{33}{16}\cdot\frac{11}{24}=\frac{121}{128}\]

2) \(4\frac{2}{9}=\frac{38}{9}\), \(3\frac{5}{14}=\frac{47}{14}\), \(6\frac{11}{21}=\frac{137}{21}\).

\[\frac{38}{9}x+\frac{47}{14}=\frac{137}{21}\]

Перенесём дробь \(\frac{47}{14}\) вправо:

\[\frac{38}{9}x=\frac{137}{21}-\frac{47}{14}=\frac{274}{42}-\frac{141}{42}=\frac{133}{42}=\frac{19}{6}\]

\[x=\frac{19}{6}\cdot\frac{9}{38}=\frac{3}{4}\]

4) Сложим коэффициенты при \(x\):

\[\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{20+15+12}{60}=\frac{47}{60}\]

Так как \(1\frac{19}{75}=\frac{94}{75}\), получаем:

\[\frac{47}{60}x=\frac{94}{75}\]

\[x=\frac{94}{75}\cdot\frac{60}{47}=\frac{8}{5}\]

Ответ: 1) \(x=\frac{121}{128}\); 2) \(x=\frac{3}{4}\); 4) \(x=\frac{8}{5}\).