a) \(\frac{3}{4} - \frac{2}{3}:x = \frac{{ - 1}}{{20}}\)

\(\frac{2}{3}:x = \frac{3}{4} - \frac{{ - 1}}{{20}}\)

\(\frac{2}{3}:x = \frac{4}{5}\)

\(x = \frac{2}{3}:\frac{4}{5}\)

\(x = \frac{5}{6}\)

Vậy \(x = \frac{5}{6}\).

b) \(\left| {2x - \frac{5}{6}} \right| - \frac{7}{3} = \frac{{ - 11}}{{18}}\)

\(\left| {2x - \frac{5}{6}} \right| = \frac{{ - 11}}{{18}} + \frac{7}{3}\)

\(\left| {2x - \frac{5}{6}} \right| = \frac{{31}}{{18}}\)

TH1: \(2x - \frac{5}{6} = \frac{{31}}{{18}}\)

\(2x = \frac{{31}}{{18}} + \frac{5}{6}\)

\(2x = \frac{{23}}{9}\)

\(x = \frac{{23}}{{18}}\).

TH2: \(2x - \frac{5}{6} = \frac{{ - 31}}{{18}}\)

\(2x = \frac{{ - 31}}{{18}} + \frac{5}{6}\)

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

\(x = \frac{{ - 4}}{9}\).

Vậy \(x \in \left\{ {\frac{{23}}{{18}};\,\,\frac{{ - 4}}{9}} \right\}\).

c) \(2{(5 - 4x)^2} - 17 = \frac{{ - 55}}{9}\)

\[2{(5 - 4x)^2} = 17 - \frac{{55}}{9}\]

\[2{(5 - 4x)^2} = \frac{{98}}{9}\]

\[{(5 - 4x)^2} = \frac{{49}}{9}\]

\[{(5 - 4x)^2} = {\left( {\frac{7}{3}} \right)^2}\]

TH1: \[5 - 4x = \frac{7}{3}\]

\[4x = 5 - \frac{7}{3}\]

\[4x = \frac{8}{3}\]

\[x = \frac{2}{3}\].

TH2: \[5 - 4x = \frac{{ - 7}}{3}\]

\[4x = 5 + \frac{7}{3}\]

\[4x = \frac{{22}}{3}\]

\[x = \frac{{11}}{6}\].

Vậy \[x \in \left\{ {\frac{2}{3};\,\,\frac{{11}}{6}} \right\}\].

Số nào sau đây là số hữu tỉ dương?