Tìm giá trị x , biết: a) 3/2 x + 2 1/4 = 8 ; b) 2/5 x + ( 3/4 )^2 = x − 1/5 ;

a) \(\frac{3}{2}x + 2\frac{1}{4} = 8\);          

\(\frac{3}{2}x + \frac{9}{4} = 8\)

\(\frac{3}{2}x = 8 - \frac{9}{4}\)

\(\frac{3}{2}x = \frac{{23}}{4}\)

\(x = \frac{{23}}{6}\).

Vậy \(x = \frac{{23}}{6}\).

b) \(\frac{2}{5}x + {\left( {\frac{3}{4}} \right)^2} = x - \frac{1}{5}\)                   

\[\frac{2}{5}x + \frac{9}{{16}} = x - \frac{1}{5}\]

\[x - \frac{2}{5}x = \frac{9}{{16}} + \frac{1}{5}\]

\[\frac{3}{5}x = \frac{{61}}{{80}}\]

\[x = \frac{{61}}{{48}}\]

Vậy \[x = \frac{{61}}{{48}}\].

c) \[3\,\,.\,\,{\left( {x - \frac{1}{8}} \right)^2} - \sqrt {\frac{{81}}{4}} = \frac{{ - 45}}{{64}}\]

\[3\,\,.\,\,{\left( {x - \frac{1}{8}} \right)^2} - \frac{9}{2} = \frac{{ - 45}}{{64}}\]

\[3\,\,.\,\,{\left( {x - \frac{1}{8}} \right)^2} = \frac{{ - 45}}{{64}} + \frac{9}{2}\]

\[3\,\,.\,\,{\left( {x - \frac{1}{8}} \right)^2} = \frac{{243}}{{64}}\]

\[{\left( {x - \frac{1}{8}} \right)^2} = \frac{{81}}{{64}}\]

TH1: \[x - \frac{1}{8} = \frac{9}{8}\]

\[x = \frac{9}{8} + \frac{1}{8}\]

\[x = \frac{5}{4}\]

TH2: \[x - \frac{1}{8} = \frac{{ - 9}}{8}\]

\[x = \frac{{ - 9}}{8} + \frac{1}{8}\]

\[x = - 1\]

Vậy \[x \in \left\{ {\frac{5}{4};\,\, - 1} \right\}\].