Đáp án:
$\max\left(\dfrac{x}{(x+2014)^2}\right)=\dfrac{1}{8056}\Leftrightarrow x = 2014$
Giải thích các bước giải:
$\quad \dfrac{x}{(x+2014)^2}\qquad (x>0)$
$=\dfrac{1}{\left(\sqrt x +\dfrac{2014}{\sqrt x}\right)^2}$
Áp dụng bất đẳng thức $AM-GM$ ta được:
$\quad \sqrt x +\dfrac{2014}{\sqrt x} \geq 2\sqrt{\sqrt x \cdot\dfrac{2014}{\sqrt x}}$
$\to \sqrt x +\dfrac{2014}{\sqrt x} \geq 2\sqrt{2014}$
$\to \left(\sqrt x +\dfrac{2014}{\sqrt x}\right)^2 \geq 4.2014 = 8056$
$\to \dfrac{1}{\left(\sqrt x +\dfrac{2014}{\sqrt x}\right)^2} \leq \dfrac{1}{8056}$
Hay $\dfrac{x}{(x+2014)^2}\leq \dfrac{1}{8056}$
Dấu $=$ xảy ra $\Leftrightarrow \sqrt x = \dfrac{2014}{\sqrt x}\Leftrightarrow x = 2014$
Vậy $\max\left(\dfrac{x}{(x+2014)^2}\right)=\dfrac{1}{8056}\Leftrightarrow x = 2014$
