ĐKXĐ: \(x\ge2\)
\(pt\Leftrightarrow x-2+x+1+2\sqrt{\left(x-2\right)\left(x+1\right)}=9\)
\(\Leftrightarrow\sqrt{x^2-x-2}=5-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}5-x\ge0\\x^2-x-2=\left(5-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le5\\9x=27\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le5\\x=3\left(tm\right)\end{matrix}\right.\)
Vậy pt đã cho có nghiệm \(x=3\)