Tìm x , biết: (a) 4/5 x + 5/6 = 41/30

a) \(\frac{4}{5}x + \frac{5}{6} = \frac{{41}}{{30}}\)

\(\frac{4}{5}x = \frac{{41}}{{30}} - \frac{5}{6}\)

\(\frac{4}{5}x = \frac{8}{{15}}\)

\(x = \frac{8}{{15}}:\frac{4}{5}\)

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

Vậy \(x = \frac{2}{3}\).

b) \(\left| {2x - \frac{7}{6}} \right| + \frac{4}{3} = \frac{{11}}{6}\)

\(\left| {2x - \frac{7}{6}} \right| = \frac{{11}}{6} - \frac{4}{3}\)

\(\left| {2x - \frac{7}{6}} \right| = \frac{1}{2}\)

TH1: \(2x - \frac{7}{6} = \frac{1}{2}\)

\(2x = \frac{1}{2} + \frac{7}{6}\)

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

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

TH2: \(2x - \frac{7}{6} = \frac{{ - 1}}{2}\)

\(2x = \frac{{ - 1}}{2} + \frac{7}{6}\)

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

\(x = \frac{1}{3}\)

Vậy \(x \in \left\{ {\frac{5}{6};\,\,\frac{1}{3}} \right\}\).

c) \(\left( {{2^x} - 16} \right)\left( {2\sqrt x - 6} \right) = 0\)

TH1: \({2^x} - 16 = 0\)

\({2^x} = 16\)

\({2^x} = {2^4}\)

\(x = 4\).

TH2: \(2\sqrt x - 6 = 0\)

\(2\sqrt x = 6\)

\(\sqrt x = 3\)

\(x = 9\).

Vậy \(x \in \{ 4;\,\,9\} \).