\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=x-1\)(dkxd
\(x\ge1\) )
\(\Leftrightarrow\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}\)
\(+\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}=x-1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(=x-1\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|=x-1\)
\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=x-1\)
TH1: \(\sqrt{x-1}\ge1\Leftrightarrow x-1\ge1\Leftrightarrow x\ge2\)(thỏa mãn điều kiện xác định )
\(\Leftrightarrow\sqrt{x-1}+1+\sqrt{x-1}-1=x-1\)
\(\Leftrightarrow2\sqrt{x-1}=x-1\)vì \(x\ge2\Leftrightarrow x-1>0\)
\(\Rightarrow4\left(x-1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow x-1=4\Leftrightarrow x=5\left(tm\right)\)
TH2:\(\sqrt{x-1}< 1\Leftrightarrow x-1< 1\Leftrightarrow x< 2\) kết hợp với điều kiện thì\(1\le x< 2\)
\(\Leftrightarrow\sqrt{x-1}+1+1-\sqrt{x-1}=x-1\)
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(ktm\right)\)
Vậy S={5}
