\(\left(2x+1\right)^2-\frac{4}{9}=0\)
\(\Leftrightarrow\left(2x+1\right)^2=\frac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x+1\right)^2=\left(\frac{2}{3}\right)^2\\\left(2x+1\right)^2=\left(-\frac{2}{3}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\frac{2}{3}\\2x+1=-\frac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-\frac{1}{3}\\2x=-\frac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{6}\\x=-\frac{5}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{6};-\frac{5}{6}\right\}\)
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