Vì \(2\ne3\) mà \(\left(3x+1\right)^2=\left(3x+1\right)^3\) nên
\(\left[{}\begin{matrix}3x+1=-1\\3x+1=0\\3x+1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=-\dfrac{1}{3}\\x=0\end{matrix}\right.\)
Vậy...........
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Ta có : \(\left(3x+1\right)^2=\left(3x+1\right)^3\)
\(9x^2+1=27x^2+1\)
\(1-1=27x^2-9x^2\)
\(0=\left(27-9\right)x^2\)
\(16x^2=0\)
\(x^2=0\)
x = \(0\)
Vậy x = 0
\(\left(3x+1\right)^2=\left(3x+1\right)^3\Leftrightarrow9x^2+6x+1=27x^3+27x^2+9x+1\)
\(\Leftrightarrow27x^3+27x^2+9x+1-9x^2-6x-1=0\)
\(\Leftrightarrow27x^3+18x^2+3x=0\Leftrightarrow x\left(27x^2+18x+3\right)=0\)
\(x\left(27x^2+9x+9x+3\right)=0\Leftrightarrow x\left(9x\left(3x+1\right)+3\left(3x+1\right)\right)\)
\(\Leftrightarrow x\left(9x+3\right)\left(3x+1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\9x+3=0\\3x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\9x=-3\\3x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\) vậy \(x=0;x=-\dfrac{1}{3}\)
