ĐKXĐ : \(x\ge1\)
Phương trình
\(\Leftrightarrow\) \(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\) \(\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\) \(\left(I\right)\)
# TH1 : \(\sqrt{x-1}-1\ge0\) \(\Leftrightarrow\) \(x\ge2\)
Khi đó \(\left(I\right)\Leftrightarrow\) \(\sqrt{x-1}+1+\sqrt{x-1}-1=2\)
\(\Leftrightarrow\) \(2\sqrt{x-1}=2\) \(\Leftrightarrow\) \(\sqrt{x-1}=1\)
\(\Leftrightarrow\) \(x-1=1\) \(\Leftrightarrow\) \(x=2\) ( thỏa mãn )
# TH2 : \(\sqrt{x-1}-1\le0\) \(\Leftrightarrow\) \(1\le x\le2\)
Khi đó (I) \(\Leftrightarrow\) \(\sqrt{x-1}+1+1-\sqrt{x-1}=2\) (luôn đúng)
Vậy pt có nghiệm \(1\le x\le2\)
