Lời giải:
Đặt \(\sqrt{x+2}=a(a\geq 0)\Rightarrow 2=a^2-x\)
Khi đó pt đã cho trở thành:
\(x^3-3x^2+2a^3-3x.2=0\)
\(\Leftrightarrow x^3-3x^2+2a^3-3x(a^2-x)=0\)
\(\Leftrightarrow x^3+2a^3-3xa^2=0\)
\(\Leftrightarrow x(x^2-a^2)-2a^2(x-a)=0\)
\(\Leftrightarrow (x-a)(x^2+xa-2a^2)=0\)
\(\Leftrightarrow (x-a)[(x^2-a^2)+a(x-a)]=0\)
\(\Leftrightarrow (x-a)^2(x+2a)=0\)
TH1: \(x-a=0\Rightarrow x=a=\sqrt{x+2}\Rightarrow \left\{\begin{matrix} a\geq 0\\ x^2=x+2\end{matrix}\right.\)
\(\Rightarrow x=2\)
TH2: \(x+2a=0\Rightarrow x=-2a=-2\sqrt{x+2}\)
\(\Rightarrow \left\{\begin{matrix} x\leq 0\\ x^2=4(x+2)\end{matrix}\right.\Rightarrow x=2-2\sqrt{3}\)
Vậy PT có nghiệm \(x\in \left\{2-2\sqrt{3}; 2\right\}\)
