Giải phương trình sau: \(\frac{3}{2} + \frac{4}{3}\left( {3x - \frac{1}{2}} \right) = \frac{1}{3}x + 2\).

Đáp án: \(\frac{7}{{22}}\)

Ta có: \(\frac{3}{2} + \frac{4}{3}\left( {3x - \frac{1}{2}} \right) = \frac{1}{3}x + 2\)

\(\frac{3}{2} + \frac{4}{3}.3x - \frac{4}{3}.\frac{1}{2} = \frac{1}{3}x + 2\)

\(\frac{3}{2} + 4x - \frac{2}{3} = \frac{1}{3}x + 2\)

\(4x - \frac{1}{3}x = 2 - \frac{3}{2} + \frac{2}{3}\)

\(\frac{{11}}{3}x = \frac{7}{6}\)

\(x = \frac{7}{{22}}\).

Vậy \(x = \frac{7}{{22}}\).