Ta có: \(4x\left(2x^2-1\right)+27=\left(4x^2+6x+9\right)\left(2x+3\right)\)
\(\Leftrightarrow8x^3-4x+27=8x^3+12x^2+12x^2+18x+18x+27\)
\(\Leftrightarrow8x^3-4x+27-8x^3-24x^2-36x-27=0\)
\(\Leftrightarrow-24x^2-40x=0\)
\(\Leftrightarrow-8x\left(3x+5\right)=0\)
mà -8≠0
nên \(\left[{}\begin{matrix}x=0\\3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-5}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{-5}{3}\right\}\)
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