Đáp án:
$\min\left(\dfrac{x}{(x+10)^2}\right) = \dfrac{1}{40} \Leftrightarrow x = 10$
Giải thích các bước giải:
$\begin{array}{l}\qquad \dfrac{x}{(x+10)^2} \qquad (x \ne -10)\\ =\dfrac{1}{\left(\sqrt x + \dfrac{10}{\sqrt x}\right)^2}\\ \text{Ta có:}\\ \quad \sqrt x + \dfrac{10}{\sqrt x} \geq 2\sqrt{\sqrt x\cdot\dfrac{10}{\sqrt x}} = 2\sqrt{10}\\ \to \left(\sqrt x + \dfrac{10}{\sqrt x}\right)^2 \geq (2\sqrt{10})^2 = 40\\ \to \dfrac{1}{\left(\sqrt x + \dfrac{10}{\sqrt x}\right)^2} \leq \dfrac{1}{40}\\ \to \dfrac{x}{(x+10)^2} \leq \dfrac{1}{40}\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow \sqrt x = \dfrac{10}{\sqrt x} \Leftrightarrow x =10\\ Vậy\,\,\min\left(\dfrac{x}{(x+10)^2}\right) = \dfrac{1}{40} \Leftrightarrow x = 10\end{array}$
