Lời giải:
ĐKXĐ: \(x\geq 1\)
\(\sqrt{x+3+4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=5\)
\(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}+\sqrt{(x-1)-6\sqrt{x-1}+9}=5\)
\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}+\sqrt{(\sqrt{x-1}-3)^2}=5\)
\(\Leftrightarrow \sqrt{x-1}+2+|\sqrt{x-1}-3|=5\)
\(\Leftrightarrow |\sqrt{x-1}-3|=3-\sqrt{x-1}=-(\sqrt{x-1}-3)\)
Do đó: \(\sqrt{x-1}-3\le 0\)
\(\Rightarrow x-1\leq 9\Rightarrow x\leq 10\)
Vậy $1\leq x\leq 10$
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