Реши уравнение: 1) 10 : 1 7/8 = 8x - 2x + x - 3x; 2) 3 2/45 x + 1 7/30 x = 6 1/9; 3) (8 2/5 : x + 3 5/9) : 4 1/6 = 1 1/3; 4) (5 1/6 - x) * 2 7/10 - 1 3/14 = 3 2/7.

Решим первое уравнение:

\[10 : 1 \frac{7}{8} = 8x - 2x + x - 3x;\]

\[10 : \frac{15}{8} = 4x;\]

\[10 \cdot \frac{8}{15} = 4x;\]

\[\frac{16}{3} = 4x;\]

\[x = \frac{16}{3} : 4;\]

\[x = \frac{16}{3} \cdot \frac{1}{4};\]

\[x = \frac{4}{3}.\]

Ошибка в условии. Верно:\[x = \frac{4}{3}.\]

Решим второе уравнение:

\[3 \frac{2}{45}x + 1 \frac{7}{30}x = 6 \frac{1}{9};\]

\[\frac{137}{45}x + \frac{37}{30}x = \frac{55}{9};\]

\[\frac{274}{90}x + \frac{111}{90}x = \frac{55}{9};\]

\[\frac{385}{90}x = \frac{55}{9};\]

\[x = \frac{55}{9} : \frac{385}{90};\]

\[x = \frac{55}{9} \cdot \frac{90}{385};\]

\[x = \frac{1}{1} \cdot \frac{10}{7};\]

\[x = \frac{10}{7}.\]

Ошибка в условии. Верно:\[x = \frac{10}{7}.\]

Решим третье уравнение:

\[\left(8 \frac{2}{5} : x + 3 \frac{5}{9}\right) : 4 \frac{1}{6} = 1 \frac{1}{3};\]

\[\left(\frac{42}{5} : x + \frac{32}{9}\right) : \frac{25}{6} = \frac{4}{3};\]

\[\frac{42}{5} : x + \frac{32}{9} = \frac{4}{3} \cdot \frac{25}{6};\]

\[\frac{42}{5} : x + \frac{32}{9} = \frac{50}{9};\]

\[\frac{42}{5} : x = \frac{50}{9} - \frac{32}{9};\]

\[\frac{42}{5} : x = \frac{18}{9};\]

\[\frac{42}{5} : x = 2;\]

\[x = \frac{42}{5} : 2;\]

\[x = \frac{42}{5} \cdot \frac{1}{2};\]

\[x = \frac{21}{5}.\]

Ошибка в условии. Верно:\[x = \frac{21}{5}.\]

Решим четвертое уравнение:

\[\left(5 \frac{1}{6} - x\right) \cdot 2 \frac{7}{10} - 1 \frac{3}{14} = 3 \frac{2}{7};\]

\[\left(\frac{31}{6} - x\right) \cdot \frac{27}{10} - \frac{17}{14} = \frac{23}{7};\]

\[\left(\frac{31}{6} - x\right) \cdot \frac{27}{10} = \frac{23}{7} + \frac{17}{14};\]

\[\left(\frac{31}{6} - x\right) \cdot \frac{27}{10} = \frac{46}{14} + \frac{17}{14};\]

\[\left(\frac{31}{6} - x\right) \cdot \frac{27}{10} = \frac{63}{14};\]

\[\left(\frac{31}{6} - x\right) \cdot \frac{27}{10} = \frac{9}{2};\]

\[\frac{31}{6} - x = \frac{9}{2} : \frac{27}{10};\]

\[\frac{31}{6} - x = \frac{9}{2} \cdot \frac{10}{27};\]

\[\frac{31}{6} - x = \frac{1}{1} \cdot \frac{5}{3};\]

\[\frac{31}{6} - x = \frac{5}{3};\]

\[-x = \frac{5}{3} - \frac{31}{6};\]

\[-x = \frac{10}{6} - \frac{31}{6};\]

\[-x = -\frac{21}{6};\]

\[x = \frac{7}{2}.\]

Ошибка в условии. Верно:\[x = \frac{7}{2}.\]