ĐXKĐ: \(x\ge2\)
Đặt \(\sqrt{x-2}=t\ge0\Rightarrow x=t^2+2\)
Pt trở thành:
\(2t^3=-\left(t^2+2\right)^2+3\left(t^2+2\right)+3\)
\(\Leftrightarrow t^4+2t^3+t^2=5\)
\(\Leftrightarrow\left(t^2+t\right)^2=5\)
\(\Leftrightarrow t^2+t=\sqrt{5}\)
\(\Leftrightarrow t=\frac{\sqrt{1+4\sqrt{5}}-1}{2}\)
\(\Leftrightarrow x=t^2+2=\frac{5+2\sqrt{5}-\sqrt{1+4\sqrt{5}}}{2}\)