(1,75 điểm) Tìm x, biết:

a) \(\frac{1}{5} - x = \frac{{ - 3}}{4}\);                         

b) \({\left( {x - \frac{2}{{15}}} \right)^3} = \frac{8}{{125}}\);                                        

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

a) \(\frac{1}{5} - x = \frac{{ - 3}}{4}\)

     \(x = \frac{1}{5} + \frac{3}{4}\)

     \(x = \frac{{19}}{{20}}\)

Vậy \(x = \frac{{19}}{{20}}\).

b) \({\left( {x - \frac{2}{{15}}} \right)^3} = \frac{8}{{125}}\)

     \({\left( {x - \frac{2}{{15}}} \right)^3} = {\left( {\frac{2}{5}} \right)^3}\)

Suy ra \(x - \frac{2}{{15}} = \frac{2}{5}\)

            \(x = \frac{2}{5} + \frac{2}{{15}}\)

            \(x = \frac{8}{{15}}\)

Vậy \(x = \frac{8}{{15}}\).

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:

\(\frac{8}{3} - 3x =  - \frac{7}{6}\)

      \(3x = \frac{8}{3} + \frac{7}{6}\)

      \(3x = \frac{{23}}{6}\)

        \(x = \frac{{23}}{6}:3\)

     \(x = \frac{{23}}{{18}}\)