ĐKXĐ: \(2x-1\ge0\Leftrightarrow x\ge\dfrac{1}{2}\)
\(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
\(\Rightarrow\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=2\)
\(\Rightarrow\sqrt{\left(2x-1\right)+2\sqrt{2x-1}+1}+\sqrt{\left(2x-1\right)-2\sqrt{2x-1}+1}=2\)
\(\Rightarrow\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
\(\Rightarrow\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|=2\)
\(\Rightarrow\sqrt{2x-1}+\left|\sqrt{2x-1}-1\right|=1\) (*)
TH1: \(\sqrt{2x-1}-1\ge0\Leftrightarrow\sqrt{2x-1}\ge1\Leftrightarrow x\ge1\)
(*) \(\Leftrightarrow2\sqrt{2x-1}-1=1\Leftrightarrow\sqrt{2x-1}=1\) \(\Leftrightarrow2x-1=1\Leftrightarrow x=1\) (TM)
TH2: \(\sqrt{2x-1}-1< 0\Leftrightarrow2x-1< 1\Leftrightarrow x< 1\)
(*) \(\Leftrightarrow\sqrt{2x-1}+1-\sqrt{2x-1}=1\Leftrightarrow1=1\) (Luôn đúng)
Kết hợp ĐKXĐ ta được tập nghiệm phương trình là: \(S=\left\{x\in R|\dfrac{1}{2}\le x\le1\right\}\)
