Tìm \(x\), biết:

c) \(\frac{7}{4} - \left( {x + \frac{5}{3}} \right) = \frac{{ - 12}}{5}\);    

b) \(\left( {1 + \frac{2}{3} - \frac{1}{4}} \right){\left( {0,8 - \frac{3}{4}} \right)^2}\)

\( = \left( {\frac{{12}}{{12}} + \frac{8}{{12}} - \frac{3}{{12}}} \right){\left( {\frac{4}{5} - \frac{3}{4}} \right)^2}\)

\( = \left( {\frac{{17}}{{12}}} \right){\left( {\frac{1}{{20}}} \right)^2}\)

\( = \frac{{17}}{{12}}\, \cdot \,\frac{1}{{400}} = \frac{{17}}{{4800}}\).

c) \[\left| {\frac{1}{3}\sqrt {x - 1} - \frac{2}{9}} \right| - \frac{1}{6} = \frac{1}{9}\]

\[\left| {\frac{1}{3}\sqrt {x - 1} - \frac{2}{9}} \right| = \frac{1}{9} + \frac{1}{6}\]

\[\left| {\frac{1}{3}\sqrt {x - 1} - \frac{2}{9}} \right| = \frac{5}{{18}}\]

TH1: \[\frac{1}{3}\sqrt {x - 1} - \frac{2}{9} = \frac{5}{{18}}\]

\[\frac{1}{3}\sqrt {x - 1} = \frac{1}{2}\]

\[\sqrt {x - 1} = \frac{3}{2}\]

\( \Rightarrow x - 1 = \frac{9}{4}\)

\(x = \frac{{13}}{4}\)

TH2: \[\frac{1}{3}\sqrt {x - 1} - \frac{2}{9} = \frac{{ - 5}}{{18}}\]

 (vô lí vì \[\frac{1}{3}\sqrt {x - 1} \ge 0;\frac{{ - 1}}{{18}} < 0\])