\(\frac{1}{\sqrt{2}}.A=\frac{\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}}{\sqrt{\left(2x-1\right)+2\sqrt{2x-1}+1}-\sqrt{\left(2x-1\right)-2\sqrt{2x-1}+1}}\)

                   \(=\frac{\sqrt{\left[\left(\sqrt{x-1}+1\right)\right]^2}+\sqrt{\left[\left(\sqrt{x-1}-1\right)^2\right]}}{\sqrt{\left[\sqrt{2x-1}+1\right]^2}-\sqrt{\left[\left(\sqrt{2x-1}\right)-1\right]^2}}\)

                     \(=\frac{\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|}{\left|\sqrt{2x-1}+1\right|-\left|\sqrt{2x-1}-1\right|}\)

DO X>2 NÊN TOÀN BỘ BIỂU THỨC TRONG TRỊ TUYỆT ĐỐI ĐỀU DƯƠNG 

\(\frac{1}{\sqrt{2}}.A=\frac{2\sqrt{x-1}}{2}=\sqrt{x-1}\)

=>\(A=\frac{\sqrt{x-1}}{\sqrt{2}}\)

\(\frac{A}{\sqrt{2}}\)=\(\frac{\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}}{\sqrt{2x-1+2\sqrt{2x-1}+1}-\sqrt{2x-1-2\sqrt{2x-1}+1}}\) (DK \(x\ge1\)

            \(=\frac{\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|}{\sqrt{2x-1}+1-\left|\sqrt{2x-1}-1\right|}\) 

vs  \(x\ge2\) \(\frac{\sqrt{x-1}+1+\sqrt{x-1}-1}{\sqrt{2x-1}+1-\sqrt{2x-1}+1}=\frac{2\sqrt{x-1}}{2}=\sqrt{x-1}\) \(\Rightarrow A=\sqrt{2x-2}\)

vs \(1\le x< 2\) \(\frac{\sqrt{x-1}+1+1-\sqrt{x-1}}{\sqrt{2x-1}+1-1+\sqrt{2x-1}}=\frac{1}{\sqrt{2x-1}}\) \(\Rightarrow A=\frac{\sqrt{2}}{\sqrt{2x-1}}\)

\(\sqrt{2X-1}\ge1\Leftrightarrow X\ge1\)NEN SUY RA THEO CACH LAM CUA TO 

THOI U AM BUSY SEE YOU AGAIN

làm zì mà dài vậy \(\sqrt{x+2\sqrt{x-1}}=\sqrt{x-1+2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}-1\right)^2}\)