Tìm x của biểu thức, biết: a) − 1/2 x + 1/2 = 2/3 ; b) (2x + 1)/ 6 = (3 − x)/ 9 ;

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

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

     \( - \frac{1}{2}x = \frac{1}{6}\)

     \(x = \frac{1}{6}:\left( { - \frac{1}{2}} \right)\)

     \(x = \frac{{ - 1}}{3}\)

Vậy \(x = \frac{{ - 1}}{3}\).

b) \(\frac{{2x + 1}}{6} = \frac{{3 - x}}{9}\)

   \(9\left( {2x + 1} \right) = 6\left( {3 - x} \right)\)

   \(18x + 9 = 18 - 6x\)

         \(24x = 9\)

             \(x = \frac{9}{{24}} = \frac{3}{8}\)

   Vậy \(x = \frac{3}{8}\).

c) \(\left| {\frac{8}{3} - 3x} \right| - \frac{2}{3} = \frac{1}{2}\)

    \(\left| {\frac{8}{3} - 3x} \right| = \frac{1}{2} + \frac{2}{3}\)

    \(\left| {\frac{8}{3} - 3x} \right| = \frac{7}{6}\)

Trường hợp 1: \(\frac{8}{3} - 3x = \frac{7}{6}\)       \(3x = \frac{8}{3} - \frac{7}{6}\)       \(3x = \frac{9}{6} = \frac{3}{2}\)         \(x = \frac{3}{2}:3\)         \(x = \frac{1}{2}\) Vậy \(x \in \left\{ {\frac{1}{2};\frac{{23}}{{18}}} \right\}\).

Trường hợp 2: \(\frac{8}{3} - 3x = - \frac{7}{6}\)       \(3x = \frac{8}{3} + \frac{7}{6}\)       \(3x = \frac{{23}}{6}\)         \(x = \frac{{23}}{6}:3\)         \(x = \frac{{23}}{{18}}\)