\(DK:x\ge1\)
\(A=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}+2019\)
\(=|\sqrt{x-1}+1|+|\sqrt{x-1}-1|+2019\)
\(=|\sqrt{x-1}+1|+|1-\sqrt{x-1}|+2019\ge|\sqrt{x-1}+1+1-\sqrt{x-1}|+2019=2021\)
Dau '=' xay ra khi \(\left(\sqrt{x-1}+1\right)\left(1-\sqrt{x-1}\right)\ge0\)
TH1:
\(\hept{\begin{cases}\sqrt{x-1}+1\ge0\\1-\sqrt{x-1}\ge0\end{cases}\Leftrightarrow x=2\left(n\right)}\)
TH2:
\(\hept{\begin{cases}\sqrt{x-1}+1\le0\\1-\sqrt{x-1}\le0\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}\le-1\\\sqrt{x-1}\ge1\end{cases}\left(l\right)}}\)
Vay \(A_{min}=2021\)khi \(x=2\)