Tìm x: a) 3/4 x + 5/2 = 23/ 8

a) \(\frac{3}{4}x + \frac{5}{2} = \frac{{23}}{8}\)

  \(\frac{3}{4}x = \frac{3}{8}\)

   \(x = \frac{1}{2}\).

Vậy \(x = \frac{1}{2}\).

b) \(\left| {2x - \frac{1}{3}} \right| = \frac{1}{2}\)

TH1: \(2x - \frac{1}{3} = \frac{1}{2}\)

\(x = \frac{5}{{12}}\)

TH2: \(2x - \frac{1}{3} =  - \frac{1}{2}\)

\(x =  - \frac{1}{{12}}\)

c) \[\left( {{3^x} - 27} \right)\left( {2\sqrt x  - 4} \right) = 0\]

TH1: \[{3^x} - 27 = 0\]

\[x = 3\]

TH2: \[2\sqrt x  - 4 = 0\]

\[x = 4\]

Vậy \[x \in \{ 3;\,\,4\} \].

(Chú ý: HS không cần đặt ĐK \[x \ge 0\]).