\(\left(2x-1\right)^3+3\left(2x+3\right)^2=26\)
\(\Leftrightarrow\left(2x\right)^3-3\left(2x\right)^2+3.2x-1+3\left(4x^2+12x+9\right)=26\)
\(\Leftrightarrow8x^3-12x^2+6x-1+12x^2+36x+27=26\)
\(\Leftrightarrow8x^3+42x+26=26\)
\(\Leftrightarrow8x^3+42x=0\)
\(\Leftrightarrow4x\left(2x^2+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\2x^2+13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x^2=-13\left(L\right)\end{matrix}\right.\)
Vậy \(x=0\)
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