\(S=\sqrt[]{a+2\sqrt[]{a-1}}+\sqrt[]{a-2\sqrt[]{a-1}}\left(1\le a\le2\right)\)
\(\Leftrightarrow S=\sqrt[]{a-1+2\sqrt[]{a-1}+1}+\sqrt[]{a-1-2\sqrt[]{a-1}+1}\)
\(\Leftrightarrow S=\sqrt[]{\left(\sqrt[]{a-1}+1\right)^2}+\sqrt[]{\left(\sqrt[]{a-1}-1\right)^2}\)
\(\Leftrightarrow S=\left|\sqrt[]{a-1}+1\right|+\left|\sqrt[]{a-1}-1\right|\left(1\right)\)
Ta lại có \(1\le a\le2\)
\(\Leftrightarrow0\le\sqrt[]{a-1}\le1\)
\(\Leftrightarrow-1\le\sqrt[]{a-1}-1\le0\)
\(\Leftrightarrow S=\sqrt[]{a-1}+1-\sqrt[]{a-1}+1\)
\(\Leftrightarrow S=2\)
\(S=\sqrt{a+2\sqrt{a-1}}+\sqrt{a-2\sqrt{a-1}}\) với \(1\le a\le2\)
Đặt \(t=\sqrt{a-1}\Leftrightarrow a=t^2+1\)
\(S=\sqrt{t^2+1+2t}+\sqrt{t^2+1-2t}\)
\(S=\sqrt{\left(t+1\right)^2}+\sqrt{\left(t-1\right)^2}\)
\(S=\left|t+1\right|+\left|t-1\right|\)
\(S= \left|\sqrt{a-1}+1\right|+\left|\sqrt{a-1}-1\right|\)
TH1: \(a=1\)
\(S=\left|1\right|+\left|-1\right|\)
\(S=2\)
TH2: \(1< a\le2\)
\(S=\sqrt{a-1}+1+\sqrt{a-1}-1\)
\(S=2\sqrt{a-1}\)