ĐKXĐ: \(x\ge\frac{1}{2}\)
\(A=\sqrt{2x-1+2\sqrt{2x-1}+1}-\sqrt{2x-1-2\sqrt{2x-1}+1}\)
\(=\sqrt{\left(\sqrt{2x-1}+1\right)^2}-\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)
\(=\sqrt{2x-1}+1-\left|\sqrt{2x-1}-1\right|\)
\(=\left[{}\begin{matrix}2\left(x\ge1\right)\\2\sqrt{2x-1}\left(\frac{1}{2}\le x< 1\right)\end{matrix}\right.\)
Để \(A< 1\Rightarrow\frac{1}{2}\le x< 1\)
\(2\sqrt{2x-1}< 1\Leftrightarrow\sqrt{2x-1}< \frac{1}{2}\Leftrightarrow2x< \frac{5}{4}\Rightarrow x< \frac{5}{8}\)
\(\Rightarrow\frac{1}{2}\le x< \frac{5}{8}\)