Bài giải
\(4x\left(2x^2-1\right)+27=\left(4x^2+6x+9\right)\left(2x+3\right)\)
\(8x^3-4x+27=8x^3+12x^2+18x+12x^2+18x+27\)
\(8x^3-4x+27=8x^3+24x^2+36x+27\)
\(8x^3-4x+27-8x^3-36x-27=24x^2\)
\(-40x=24x^2\)
\(\frac{3}{5}x^2=x\)
\(\frac{3}{5}x^2-x=0\)
\(x\left(\frac{3}{5}x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\\frac{3}{5}x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\frac{3}{5}x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{5}{3}\end{matrix}\right.\)
\(\Rightarrow\text{ }x\in\left\{0\text{ ; }\frac{5}{3}\right\}\)
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