`1/9 - (2x + 1/3)^2 = 0`

`⇔ (2x + 1/3)^2 = 1/9`

`⇔` \(\left[ \begin{array}{l}(2x+\frac{1}{3})^2 = (\frac{1}{3})^2\\(2x+\frac{1}{3})^2 = (- \frac{1}{3})^2\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}2x+\frac{1}{3} = \frac{1}{3}\\2x+\frac{1}{3} = - \frac{1}{3}\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}2x= \frac{1}{3} - \frac{1}{3} \\2x = - \frac{1}{3} - \frac{1}{3} \end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}2x= 0 \\2x = - \frac{2}{3}  \end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x = 0 \\x = - \frac{1}{3}  \end{array} \right.\) 

Vậy `x=0` hoặc `x=-1/3`