\(\left(x+2\right)=\left(-2x-1\right)^3\)
\(\Leftrightarrow x+2=-\left(8x^3+12x^2+6x+1\right)\)
\(\Leftrightarrow8x^3+12x^2+7x+3=0\)
\(\Leftrightarrow\left(8x^3+8x^2\right)+\left(4x^2+4x\right)+\left(3x+3\right)=0\)
\(\Leftrightarrow8x^2\left(x+1\right)+4x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(8x^2+4x+3\right)\left(x+1\right)=0\)
\(\Leftrightarrow x+1=0\) (vì \(8x^2+4x+3=8\left(x^2+\dfrac{1}{2}x+\dfrac{3}{8}\right)=8\left[\left(x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}\right)+\dfrac{5}{16}\right]=8\left(x+\dfrac{1}{4}\right)^2+\dfrac{5}{2}>0\))
\(\Leftrightarrow x=-1\)
Vậy \(x=-1\)
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