Question

Tìm Min

A=\(\dfrac{\sqrt{x-2018}}{x}\)+\(\dfrac{\sqrt{y-2019}}{y}\)

Answer 1

điều kiện xác định : \(x\ge2018;y\ge2019\)

\(\Rightarrow A=\dfrac{\sqrt{x-2018}}{x}+\dfrac{\sqrt{y-2019}}{y}\ge0\)

\(\Rightarrow A_{min}=0\) khi \(\left\{{}\begin{matrix}x-2018=0\\x-2019=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2018\\x=2019\end{matrix}\right.\)

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