Ta có: \(27x^2\left(x+3\right)=12\left(x^2+3x\right)\)
\(\Leftrightarrow27x^3+81x^2=12x^2+36x\)
\(\Leftrightarrow27x^3+81x^2-12x^2-36x=0\)
\(\Leftrightarrow27x^3+69x^2-36x=0\)
\(\Leftrightarrow3x\left(9x^2+23x-12\right)=0\)
Vì 3>0
nên \(x\left(9x^2+23x-12\right)=0\)
\(\Leftrightarrow x\left(9x^2+27x-4x-12\right)=0\)
\(\Leftrightarrow x\left[9x\left(x+3\right)-4\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(9x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\9x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\9x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\frac{4}{9}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-3;\frac{4}{9}\right\}\)
