Question

Tìm MAX :

\(\sqrt{x-2}+2\sqrt{x+1}+2019-x\)

Answer 1

\(Đk:x\ge2\)

Đặt \(A=\sqrt{x-2}+2\sqrt{x+1}+2019-x\)

\(2A=2\sqrt{x-2}+4\sqrt{x+1}+4038-2x\)

\(=\left[\left(-x+2\right)+2\sqrt{x-2}-1\right]+\left[\left(-x-1\right)+4\sqrt{x+1}-4\right]+4042\)

\(=-\left[\left(x-2\right)-2\sqrt{x-2}+1\right]-\left[\left(x+1\right)-4\sqrt{x+1}+4\right]+4042\)

\(=-\left(\sqrt{x-2}-1\right)^2-\left(\sqrt{x+1}-2\right)^2+4042\le4042\)

\(\Rightarrow A\le2021\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x-2}-1=0\\\sqrt{x+1}-2=0\end{matrix}\right.\Leftrightarrow x=3\)

Vậy \(Max\) của biểu thức trên là 2021, đạt tại x=3.