1.153. Решите уравнение: а) \(x+2\frac{3}{16}=3\frac{1}{8}\); б) \(x-1\frac{3}{4}=2\frac{11}{12}\); в) \(\frac{2}{9}:x=\frac{7}{18}\); г) \(\frac{3}{5}\cdot x=\frac{6}{7}\); д) \(\left(1\frac{2}{3}+\frac{4}{9}\right):x=\frac{9}{38}\); е) \(\left(8\frac{1}{2}-7\frac{1}{4}\right)\cdot x=\frac{5}{12}\).

а) \(x+2\frac{3}{16}=3\frac{1}{8}\)

\(x=3\frac{1}{8}-2\frac{3}{16}=\frac{50}{16}-\frac{35}{16}=\frac{15}{16}\).

б) \(x-1\frac{3}{4}=2\frac{11}{12}\)

\(x=2\frac{11}{12}+1\frac{3}{4}=\frac{35}{12}+\frac{21}{12}=\frac{56}{12}=4\frac{2}{3}\).

в) \(\frac{2}{9}:x=\frac{7}{18}\)

\(x=\frac{2}{9}:\frac{7}{18}=\frac{2}{9}\cdot\frac{18}{7}=\frac{4}{7}\).

г) \(\frac{3}{5}\cdot x=\frac{6}{7}\)

\(x=\frac{6}{7}:\frac{3}{5}=\frac{6}{7}\cdot\frac{5}{3}=\frac{10}{7}=1\frac{3}{7}\).

д) \(\left(1\frac{2}{3}+\frac{4}{9}\right):x=\frac{9}{38}\)

\(1\frac{2}{3}+\frac{4}{9}=\frac{5}{3}+\frac{4}{9}=\frac{19}{9}\).

\(x=\frac{19}{9}:\frac{9}{38}=\frac{19}{9}\cdot\frac{38}{9}=\frac{722}{81}=8\frac{74}{81}\).

е) \(\left(8\frac{1}{2}-7\frac{1}{4}\right)\cdot x=\frac{5}{12}\)

\(8\frac{1}{2}-7\frac{1}{4}=\frac{17}{2}-\frac{29}{4}=\frac{34}{4}-\frac{29}{4}=\frac{5}{4}\).

\(x=\frac{5}{12}:\frac{5}{4}=\frac{5}{12}\cdot\frac{4}{5}=\frac{1}{3}\).

Ответ: а) \(\frac{15}{16}\); б) \(4\frac{2}{3}\); в) \(\frac{4}{7}\); г) \(\frac{10}{7}\); д) \(\frac{722}{81}\); е) \(\frac{1}{3}\).