ĐKXĐ: \(x\ge2\)
\(x^2-6x+9+x-2-2\sqrt{x-2}+1=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x-2}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{x-2}-1=0\end{matrix}\right.\) \(\Rightarrow x=3\)
Pt có nghiệm duy nhất \(x=3\)
\(x^2-5x+8=2\sqrt{x-2}\)
\(\Leftrightarrow\left(x^2-6x+9\right)+\left(x-2-2\sqrt{x-2}+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x-2}-1\right)^2=0\)
\(\Leftrightarrow x=3\)