(1,0 điểm) Tìm x, biết:

a) \(1\frac{2}{3}.x - \frac{1}{4} = \frac{5}{6}\);                                    

b) \({\left( {5x + \frac{1}{5}} \right)^2} = \frac{4}{{25}}\).

a) \(1\frac{2}{3}.x - \frac{1}{4} = \frac{5}{6}\)

     \(\frac{5}{3}x = \frac{5}{6} + \frac{1}{4}\)

     \(\frac{5}{3}x = \frac{{10}}{{12}} + \frac{3}{{12}} = \frac{{13}}{{12}}\)

      \(x = \frac{{13}}{{12}}:\frac{5}{3}\)

      \(x = \frac{{13}}{{12}}.\frac{3}{5} = \frac{{13}}{{20}}\)

Vậy \(x = \frac{{13}}{{20}}\).

b) \({\left( {5x + \frac{1}{5}} \right)^2} = \frac{4}{{25}}\)

     \({\left( {5x + \frac{1}{5}} \right)^2} = {\left( {\frac{2}{5}} \right)^2} = {\left( { - \frac{2}{5}} \right)^2}\)

Trường hợp 1:

\(5x + \frac{1}{5} = \frac{2}{5}\)

\(5x = \frac{2}{5} - \frac{1}{5} = \frac{1}{5}\)

\(x = \frac{1}{5}:5 = \frac{1}{{25}}\)

Vậy \(x \in \left\{ {\frac{1}{{25}};\frac{{ - 3}}{{25}}} \right\}\).

Trường hợp 2:

\(5x + \frac{1}{5} = - \frac{2}{5}\)

\(5x = - \frac{2}{5} - \frac{1}{5} = \frac{{ - 3}}{5}\)

\(x = \frac{{ - 3}}{5}:5 = \frac{{ - 3}}{{25}}\)