Tìm x , biết: a) 2. [ ( 2/5 − x ) + 3x ] = − 16/5 ;

a) \(2.\left[ {\left( {\frac{2}{5} - x} \right) + 3x} \right] = - \frac{{16}}{5}\)

\(\left( {\frac{2}{5} - x} \right) + 3x = - \frac{{16}}{5}:2\)

\(\frac{2}{5} - x + 3x = - \frac{8}{5}\)

\(3x - x = - \frac{8}{5} - \frac{2}{5}\)

\(2x = - 2\)

\(x = - 1\)

Vậy \(x = - 1\).

b) \(\left| {x - \frac{1}{2}} \right| \cdot \sqrt {4\frac{9}{4}} - 2\frac{1}{4} = {\left( { - \frac{1}{2}} \right)^2}\)

\(\left| {x - \frac{1}{2}} \right| \cdot \sqrt {\frac{{25}}{4}} - \frac{9}{4} = \frac{1}{4}\)

\(\left| {x - \frac{1}{2}} \right| \cdot \frac{5}{2} = \frac{5}{2}\)

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

+) TH1: \(x - \frac{1}{2} = 1\)

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

\(x = \frac{3}{2}.\)

+) TH2: \(x - \frac{1}{2} = - 1\)

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

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

Vậy \(x \in \left\{ {\frac{3}{2};\,\,\frac{{ - 1}}{2}} \right\}\).