ĐKXĐ: \(x\ge\frac{2}{3}\)
\(\Leftrightarrow\sqrt{x+2}-2+x-\sqrt{3x-2}+x^2-2x\le0\)
\(\Leftrightarrow\frac{x-2}{\sqrt{x+2}+2}+\frac{x^2-3x+2}{x+\sqrt{3x-2}}+x\left(x-2\right)\le0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{1}{\sqrt{x+2}+2}+\frac{x-1}{x+\sqrt{3x-2}}+x\right)\le0\)
\(\Leftrightarrow x-2\le0\Rightarrow x\le2\)
Vậy nghiệm của BPT là: \(\frac{2}{3}\le x\le2\)