\(A=\sqrt{x+\sqrt{2x-1}}-\sqrt{x+\sqrt{2x+1}}\) (ĐKXĐ: \(x\ge\dfrac{1}{2}\))
\(=\sqrt{\dfrac{2x+2\sqrt{2x-1}}{2}}-\sqrt{\dfrac{2x-2\sqrt{2x-1}}{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{2x-1}\right)^2+2\sqrt{2x-1}+1^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{2x-1}\right)^2-2\sqrt{2x-1}+1^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{2x-1}+1\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{2x-1}-1\right)^2}}{\sqrt{2}}\)
\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{2x-1}+1\right|-\left|\sqrt{2x-1}-1\right|\right)\)
TH1: \(\sqrt{2x-1}-1\ge0\Leftrightarrow\sqrt{2x-1}\ge1\Leftrightarrow2x-1\ge1\Leftrightarrow x\ge1\)
\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{2x-1}+1-\sqrt{2x-1}+1\right)=\sqrt{2}\)
TH2: \(\sqrt{2x-1}-1< 0\Leftrightarrow\sqrt{2x-1}< 1\Leftrightarrow2x-1< 1\Leftrightarrow\dfrac{1}{2}\le x< 1\)
\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{2x-1}+1+\sqrt{2x-1}-1\right)=\sqrt{4x-2}\)