ĐKXĐ: \(x\ge1\)
Đặt \(\sqrt{x-1}=t\ge0\) \(\Rightarrow x=t^2+1\)
Pt trở thành:
\(\sqrt{t^2+1-2t}-t=1\)
\(\Leftrightarrow\sqrt{t^2-2t+1}=t+1\)
\(\Leftrightarrow t^2-2t+1=t^2+2t+1\)
\(\Leftrightarrow4t=0\)
\(\Rightarrow t=0\)
\(\Rightarrow\sqrt{x-1}=0\)
\(\Rightarrow x=1\)
ĐK: x≥1
Ta có: \(\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\)
\(\Leftrightarrow\left|\sqrt{x-1}-1\right|=1+\sqrt{x-1}\)
Nếu \(\sqrt{x-1}-1\ge0\Leftrightarrow x\ge2\) \(\Leftrightarrow\sqrt{x-1}-1=1+\sqrt{x-1}\)
\(\Leftrightarrow-1=1\) (vô lí)
Nếu \(\sqrt{x-1}-1< 0\Leftrightarrow x< 2\) \(\Leftrightarrow1-\sqrt{x-1}=1+\sqrt{x-1}\)
\(\Leftrightarrow2\sqrt{x-1}=0\)
\(\Leftrightarrow\sqrt{x-1}=0\)
\(\Leftrightarrow x=1\left(tm\right)\)
