ĐKXĐ: ...
\(\Leftrightarrow3x^3-9x^2+4+2\sqrt{x^3-3x^2+3}=0\)
Đặt \(\sqrt{x^3-3x^2+3}=t\ge0\Rightarrow x^3-3x^2=t^2-3\)
Pt trở thành:
\(3\left(t^2-3\right)+4+2t=0\)
\(\Leftrightarrow3t^2+2t-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-\frac{5}{3}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x^3-3x^2+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-2x-2\right)=0\)