Đáp án: $\dfrac{500}{27}$
Giải thích các bước giải:
Ta có:
$x\in[0,5]\to 0\le x, 5-x\ge 0$
$\to \dfrac{x}{2}+\dfrac{x}{2}+(5-x)\ge 3\sqrt[3]{\dfrac{x}{2}\cdot\dfrac{x}{2}\cdot(5-x)}$
$\to 5\ge 3\sqrt[3]{\dfrac{x^2(5-x)}{4}}$
$\to \dfrac{x^2(5-x)}{4}\le \dfrac{125}{27}$
$\to x^2(5-x)\le \dfrac{500}{27}$
Dấu = xảy ra khi $\dfrac{x}{2}=5-x\to x=\dfrac{10}{3}$