Question

\(\sqrt{x^2+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)

Answer 1

sửa đề chút nha :)

điều kiện xác định : \(x\ge1\)

ta có : \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2=2\left(x\ge2\right)\\2\sqrt{x-1}=2\left(1\le x< 2\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x=2\end{matrix}\right.\Rightarrow x\ge2\)

vậy \(x\ge2\)