Question

Tìm GTLN:

\(\frac{\sqrt{x-2019}}{x+2}+\frac{\sqrt{x-2020}}{x}\)

Answer 1

ĐKXĐ: \(x\ge2020\)

- Với \(x=2020\Rightarrow A=\frac{1}{2022}\)

- Với \(x>2020\)

\(A=\frac{\sqrt{x-2019}}{x-2019+2021}+\frac{\sqrt{x-2020}}{x-2020+2020}\)

\(A=\frac{1}{\sqrt{x-2019}+\frac{2021}{\sqrt{x-2019}}}+\frac{1}{\sqrt{x-2020}+\frac{2020}{\sqrt{x-2020}}}\)

\(A\le\frac{1}{2\sqrt{2021}}+\frac{1}{2\sqrt{2020}}\)

So sánh với \(\frac{1}{2022}\Rightarrow A_{max}=\frac{1}{2\sqrt{2019}}+\frac{1}{2\sqrt{2020}}\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}x-2019=2021\\x-2020=2020\end{matrix}\right.\) \(\Rightarrow x=4040\)