\(x^3+6x^2+12x+8+2\sqrt{\left(x+2\right)^3}+1-\left(9x^2+18x+9\right)=0\)
\(\Leftrightarrow\left(x+2\right)^3+2\sqrt{\left(x+2\right)^3}+1=\left(3x+3\right)^2\)
\(\Leftrightarrow\left(\sqrt{\left(x+2\right)^3}+1\right)^2=\left(3x+3\right)^2\Rightarrow\left[{}\begin{matrix}\sqrt{\left(x+2\right)^3}+1=3x+3\\\sqrt{\left(x+2\right)^2}+1=-3x-3\end{matrix}\right.\)
TH1: \(x\ge-1\Rightarrow\sqrt{\left(x+2\right)^3}+1=3x+3\Leftrightarrow\sqrt{\left(x+2\right)^3}=3x+2\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2\ge0\\\left(x+2\right)^3=\left(3x+2\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{2}{3}\\x^3-3x^2+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{2}{3}\\\left(x-2\right)^2\left(x+1\right)=0\end{matrix}\right.\)
\(\Rightarrow x=2\)
TH2: \(x< -1\Rightarrow\sqrt{\left(x+2\right)^3}+1=-3x-3\Leftrightarrow\left\{{}\begin{matrix}-3x-4\ge0\\\left(x+2\right)^3=\left(-3x-4\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{4}{3}\\x^3-3x^2-12x-8=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{4}{3}\\\left(x+1\right)\left(x^2-4x-8\right)=0\end{matrix}\right.\)
\(\Rightarrow x=2-2\sqrt{3}\)
Vậy pt đã cho có 2 nghiệm: \(\left[{}\begin{matrix}x=2\\x=2-2\sqrt{3}\end{matrix}\right.\)
