\(ĐK:x\ge1\)
\(\sqrt{x+2\sqrt{x-1}}=2\)
\(\Leftrightarrow\left(\sqrt{x+2\sqrt{x-1}}\right)^2=2^2\)
\(\Leftrightarrow x+2\sqrt{x-1}=4\)
\(\Leftrightarrow x-4+2\sqrt{x-1}=0\)
Đặt \(x-1=a;a\ge0\)
\(\Leftrightarrow a+2\sqrt{a}-3=0\)
\(\Leftrightarrow\left(\sqrt{a}-1\right)\left(\sqrt{a}+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\left(tm\right)\\\sqrt{a}=-3\left(vô.lý\right)\end{matrix}\right.\)
\(\rightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
Vậy \(S=\left\{2\right\}\)
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