Tìm \(x\) biết: \({x^2} + \dfrac{1}{9} = \dfrac{5}{3}:3\)

A. \(x \in \left\{ {\dfrac{4}{9}; - \dfrac{4}{9}} \right\}\)

B. \(x \in \left\{ {\dfrac{2}{3}; - \dfrac{2}{3}} \right\}\)

C. \(x = \dfrac{4}{9}\)

D. \(x = \dfrac{2}{3}\)

 

\({x^2} + \dfrac{1}{9} = \dfrac{5}{3}:3\)

\(\begin{array}{l}{x^2} + \dfrac{1}{9} = \dfrac{5}{3}.\dfrac{1}{3} = \dfrac{5}{9}\\{x^2} = \dfrac{5}{9} - \dfrac{1}{9}\\{x^2} = \dfrac{4}{9}\\{x^2} = {\left( {\dfrac{2}{3}} \right)^2} = {\left( { - \dfrac{2}{3}} \right)^2}\end{array}\)

\( \Rightarrow x = \dfrac{2}{3}\) hoặc \(x =  - \dfrac{2}{3}\)

Vậy \(x \in \left\{ {\dfrac{2}{3}; - \dfrac{2}{3}} \right\}\)

Chọn B.