Sửa lại tí:\(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
ĐK: \(x\ge1\)
\(\Leftrightarrow\sqrt{2}\sqrt{x+\sqrt{2x-1}}+\sqrt{2}\sqrt{x-\sqrt{2x-1}}=2\)
\(\Leftrightarrow\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=2\)
\(\Leftrightarrow\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
Vì \(x\ge1\Rightarrow\sqrt{2x-1}-1\ge\sqrt{2-1}-1=0\Rightarrow\left|\sqrt{2x-1}-1\right|=\sqrt{2x-1}-1\)
\(\Leftrightarrow\sqrt{2x-1}+1+\sqrt{2x-1}-1=2\)
\(\Leftrightarrow2\sqrt{2x-1}=2\)
\(\Leftrightarrow\sqrt{2x-1}=1\)
\(\Leftrightarrow2x-1=1\)
\(\Leftrightarrow2x=2\)
\(\Leftrightarrow x=1\) (thỏa)
Vậy S = {1}
