\(\left(3x+1\right)^2=3x+1\)
\(\Leftrightarrow\left(3x+1\right)^2-\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x+1-1\right)=0\)
\(\Leftrightarrow3x\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{3}\end{matrix}\right.\)
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