\(A=\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}\)
\(=\sqrt{a-2+2\sqrt{a-2}.2+4}+\sqrt{a-2-2\sqrt{a-2}.2+4}\)
\(=\sqrt{\left(\sqrt{a-2}\right)^2+2\sqrt{a-2}.2+2^2}+\sqrt{\left(\sqrt{a-2}\right)^2-2\sqrt{a-2}.2+2^2}\)
\(=\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}\)
\(=\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|\)
Ta có:
\(A=\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}\)
\(A=\sqrt{\left(\sqrt{a-2}\right)^2+2.2\sqrt{a-2}+2^2}+\sqrt{\left(\sqrt{a-2}\right)^2-2.2\sqrt{a-2}+2^2}\)
\(A=\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}\)
\(A=|\sqrt{a-2}+2|+|\sqrt{a-2}-2|\) (1)
=> Điều kiên: a - 2 >= 0 <=> a >= 2
(1) => \(A=\sqrt{a-2}+2+|\sqrt{a-2}-2|\)(Do số hạng đầu luôn lớn hơn 0 nên bỏ trị tuyệt đối)
TH1: \(\sqrt{a-2}-2\ge0\Rightarrow A=\sqrt{a-2}+2+\sqrt{a-2}-2\)
\(\sqrt{a-2}\ge2\Rightarrow A=2\sqrt{a-2}\)
\(a\ge6\Rightarrow A=2\sqrt{a-2}\)
(nhận)
TH2: \(\sqrt{a-2}-2\le0\Rightarrow A=\sqrt{a-2}+2-\sqrt{a-2}+2\)
\(\sqrt{a-2}\le2\Rightarrow A=4\)
\(2\le a\le6\Rightarrow A=4\) (Do đkxđ)
Vậy....
\(A=\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}\left(a\ge2\right)\)
\(A=\sqrt{\left(a-2\right)+4\sqrt{a-2}+4}+\sqrt{\left(a-2\right)-4\sqrt{a-2}+4}\)
\(A=\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}\)
\(A=\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|\)
\(A=\sqrt{a-2}+2+\left|\sqrt{a-2}-2\right|\)
\(A=\hept{\begin{cases}\sqrt{a-2}+2+\sqrt{a-2}-2\Leftrightarrow\sqrt{a-2}-2\ge0\\\sqrt{a+2}+2-\sqrt{a-2}+2\Leftrightarrow\sqrt{a-2}< 0\end{cases}}\)
\(A=\hept{\begin{cases}2\sqrt{a-2}\Leftrightarrow a-2\ge4\\4\Leftrightarrow a-2< 4\end{cases}}\)
\(A=\hept{\begin{cases}2\sqrt{a-2}\Leftrightarrow a\ge6\\4\Leftrightarrow a< 6\end{cases}}\)
Vậy \(A=\hept{\begin{cases}2\sqrt{a-2}\Leftrightarrow a\ge6\\4\Leftrightarrow2\le a\le6\end{cases}}\)
\(\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|\) (ĐKXĐ: \(a-2\ge0\Rightarrow a\ge2\) )
Vì \(\sqrt{a-2}+2\) luôn không âm
Nên \(\left|\sqrt{a-2}-2\right|\) khi bỏ dấu GTTĐ sẽ xảy ra hai trường hợp
-TH1: \(\sqrt{a-2}+2+\left|\sqrt{a-2}-2\right|=\sqrt{a-2}+2+\sqrt{a-2}-2=2\sqrt{a-2}\) (nhận)
-TH2: \(\sqrt{a-2}+2+\left|\sqrt{a-2}-2\right|=\sqrt{a-2}+2+2-\sqrt{a-2}=4\) (loại)
Vậy \(A=\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}=2\sqrt{a-2}\)
