ĐKXĐ: \(x\ge2\)
\(\sqrt{x-2-4\sqrt{x-2}+4}+\sqrt{x-2-6\sqrt{x-2}+9}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-2}-2\right)^2}+\sqrt{\left(\sqrt{x-2}-3\right)^2}=5\)
\(\Leftrightarrow\left|\sqrt{x-2}-2\right|+\left|\sqrt{x-2}-3\right|=5\)
- TH1:
\(2\le x\le6\)
\(\Rightarrow2-\sqrt{x-2}+3-\sqrt{x-2}=5\)
\(\Leftrightarrow\sqrt{x-2}=0\Rightarrow x=2\)
TH2: \(6\le x\le11\)
\(\Rightarrow\sqrt{x-2}-2+3-\sqrt{x-2}=5\Leftrightarrow1=5\) (vô nghiệm)
TH3: \(x>11\)
\(\Rightarrow\sqrt{x-2}-2+\sqrt{x-2}-3=5\)
\(\Leftrightarrow2\sqrt{x-2}=10\Rightarrow\sqrt{x-2}=5\)
\(\Rightarrow x=27\)
Vậy \(\left[{}\begin{matrix}x=2\\x=27\end{matrix}\right.\)