\(A=\sqrt{x-2}+2\sqrt{x+1}+2019-x\)
\(=\sqrt{x-2}+2\sqrt{\frac{x+1}{2}.2}+2019-x\)
\(\le\sqrt{x-2}+\frac{x+1}{2}+2+2019-x\)
\(=\frac{-x+1}{2}+\sqrt{x-2}+2021\)
\(=\frac{-x+2}{2}+2.\frac{\sqrt{x-2}}{2}-\frac{1}{2}+2021\)
\(=-\left(\frac{x-2}{2}\right)+2.\frac{\sqrt{x-2}}{\sqrt{2}}.\frac{1}{\sqrt{2}}-\frac{1}{2}+2021\)
\(=-\left(\frac{\sqrt{x-2}}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)^2+2021\)
\(\le2021\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\frac{x+1}{2}=2\\\frac{\sqrt{x-2}}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0\end{matrix}\right.\)\(\Leftrightarrow x=3\)
Vậy MinA=2021 khi và chỉ khi x=3
