Giá trị x thỏa mãn \(27{x^2} - 3 = 0\)là:

A. \(x = \dfrac{1}{9}\);

B. \(x = \dfrac{1}{3}\);

C. \(x = 3\)

D. \(x =  \pm \dfrac{1}{3}\).

\(\begin{array}{l}27{x^2} - 3 = 0\\ \Leftrightarrow 27{x^2} = 3\\ \Leftrightarrow {x^2} = \dfrac{3}{{27}} = \dfrac{1}{9}\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{1}{3}}\\{x = \dfrac{{ - 1}}{3}}\end{array}} \right.\end{array}\)

Vậy \(x \in \left\{ {\dfrac{1}{3};\dfrac{{ - 1}}{3}} \right\}\)

Chọn D