Tìm \(x,\) biết:   

a) \(\frac{5}{2} - x = \frac{{27}}{8}\).                   b) \(\frac{{ - 1}}{3} + \frac{4}{{10}}x = 0,2\).                                       c) \({\left( {3x + 2} \right)^2} = \frac{{25}}{{49}}\).

a \(\frac{5}{2} - x = \frac{{27}}{8}\)

    \(x = \frac{5}{2} - \frac{{27}}{8}\)

    \(x = \frac{{20}}{8} - \frac{{27}}{8}\)              

     \(x = \frac{{ - 7}}{8}\)

Vậy \(x = \frac{{ - 7}}{8}.\)

 b \(\frac{{ - 1}}{3} + \frac{4}{{10}}x = 0,2\)

    \(\frac{4}{{10}}x = 0,2 - \left( {\frac{{ - 1}}{3}} \right)\)

    \(\frac{4}{{10}}x = \frac{8}{{15}}\)

    \(x = \frac{8}{{15}}:\frac{4}{{10}}\)

    \(x = \frac{4}{3}\)

Vậy \(x = \frac{4}{3}\)

 c \({\left( {3x + 2} \right)^2} = \frac{{25}}{{49}}\)

\[{\left( {3x + 2} \right)^2} = {\left( {\frac{5}{7}} \right)^2} = {\left( {\frac{{ - 5}}{7}} \right)^2}\]

Trường hợp 1: \[3x + 2 = \frac{5}{7}\]

\[3x = \frac{5}{7} - 2\]

\[3x = \frac{{ - 9}}{7}\]

\[x = \frac{{ - 9}}{7}:3\]

\[x = \frac{{ - 3}}{7}\]

Trường hợp 2: \[3x + 2 = \frac{{ - 5}}{7}\]

\[3x = \frac{{ - 5}}{7} - 2\]

\[3x = \frac{{ - 19}}{7}\]

\[x = \frac{{ - 19}}{7}:3\]

\[x = \frac{{ - 19}}{{21}}\]

Vậy \[x \in \left\{ {\frac{{ - 3}}{7};\frac{{ - 19}}{{21}}} \right\}.\]