Đáp án:
$\max\left(\dfrac{x}{(x+2021)^2}\right) = \dfrac{1}{8084}\Leftrightarrow x = 2021$
Giải thích các bước giải:
$\dfrac{x}{(x+2021)^2}\qquad (x > 0)$
$=\dfrac{1}{\dfrac{1}{x}(x+2021)^2}$
$=\dfrac{1}{\left(\dfrac{x +2021}{\sqrt x}\right)^2}$
$=\dfrac{1}{\left(\sqrt x +\dfrac{2021}{\sqrt x}\right)^2}$
Ta có:
$\sqrt x +\dfrac{2021}{\sqrt x} \geq 2\sqrt{\sqrt x\cdot\dfrac{2021}{\sqrt x}}=2\sqrt{2021}$
$\to \left(\sqrt x +\dfrac{2021}{\sqrt x}\right)^2 \geq 4.2021 =8084$
$\to \dfrac{1}{\left(\sqrt x +\dfrac{2021}{\sqrt x}\right)^2}\leq \dfrac{1}{8084}$
Dấu $=$ xảy ra $\Leftrightarrow \sqrt x =\dfrac{2021}{\sqrt x}\Leftrightarrow x = 2021$
Vậy $\max\left(\dfrac{x}{(x+2021)^2}\right) = \dfrac{1}{8084}\Leftrightarrow x = 2021$
