\(\left(3x+1\right)^2=\left(3x+1\right)^4\)
\(9x^2+1=81x^4+1\)
\(1-1=81x^4-9x^2\)
\(0=9.9.x^4-9x^2\)
\(0=9.\left(9x^4-x^2\right)\)
Còn lại thì mình không biết 
P\S : hình như mình làm sai rồi thì phải 
\(\left(3x+1\right)^2=\left(3x+1\right)^4\)
\(\Rightarrow\left(3x+1\right)^2-\left(3x+1\right)^4=0\)
\(\Rightarrow\left(3x+1\right)^2\left[1-\left(3x+1\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(3x+1\right)^2=0\\1-\left(3x+1\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x+1=0\\\left(3x+1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=-1\\3x+1=1\\3x+1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\3x=0\Rightarrow x=0\\3x=-2\Rightarrow x=\dfrac{-2}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-1}{3},0,\dfrac{-2}{3}\right\}\)
