Tìm x , biết: c) trị tuyệt đối của 1/ 3 √ x − 1 − 2/ 9 − 1 /6 = 1 /9 ;

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

Vậy \(x = \frac{{13}}{4}\).