Đặt A = \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\) (\(x\ge2\)) ( \(A\ge0\))
=> A2 = \(x+2\sqrt{2x-4}+x-2\sqrt{2x-4}+2\sqrt{x+2\sqrt{2x-4}}.\sqrt{x-2\sqrt{2x-4}}\)
<=> A2 = 2x + \(2\sqrt{x^2-4\left(2x-4\right)}\)
<=> A2 = \(2x+2\sqrt{\left(x-4\right)^2}\) = \(2x+2\left|x-4\right|\)
TH1: \(x\ge4\)
<=> A2 = 2x + 2(x-4) = 4x-8
<=> A = \(\sqrt{4\left(x-2\right)}=2\sqrt{x-2}\)
TH2: \(2\le x< 4\)
<=> A2 = 2x + 2 (4-x) = 8
<=> A = \(2\sqrt{2}\)
\(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\) (đk: \(x\ge2\))
\(=\sqrt{\left(x-2\right)+2.\sqrt{2}\sqrt{x-2}+2}+\sqrt{\left(x-2\right)-2\sqrt{2}\sqrt{x-2}+2}\)
\(=\sqrt{\left(\sqrt{x-2}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{x-2}-\sqrt{2}\right)^2}\)
\(=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|\)
\(=\left[{}\begin{matrix}\sqrt{x-2}+\sqrt{2}+\sqrt{x-2}-\sqrt{2};\sqrt{x-2}\ge2\\\sqrt{x-2}+\sqrt{2}-\left(\sqrt{x-2}-\sqrt{2}\right);\sqrt{x-2}< 2\end{matrix}\right.\)
\(=\left[{}\begin{matrix}2\sqrt{x-2};x\ge6\\2\sqrt{2};2\le x< 6\end{matrix}\right.\)
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