Đáp án:
\(\max K = \dfrac{1}{8084} \Leftrightarrow x = 2021\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad K = \dfrac{x}{(x + 20121)^2}\qquad (x >0)\\
\to K = \dfrac{1}{\left(\sqrt x + \dfrac{2021}{\sqrt x}\right)^2}\\
\to K \leqslant \dfrac{1}{\left(2\sqrt{\sqrt x\cdot \dfrac{2021}{\sqrt x}}\right)^2}\qquad (BDT\ AM-GM)\\
\to K \leqslant \dfrac{1}{4\cdot 2021}\\
\to K \leqslant \dfrac{1}{8084}\\
\text{Dấu = xảy ra}\ \Leftrightarrow \sqrt x = \dfrac{2021}{\sqrt x} \Leftrightarrow x = 20201\\
\text{Vậy}\ \max K = \dfrac{1}{8084} \Leftrightarrow x = 2021
\end{array}\)
