d) $y = (2x +5)(5 - x)\qquad \left(-\dfrac{5}{2}\leq x \leq 5\right)$
$\to y = \dfrac{1}{2}(2x + 5)(10 - 2x) \leq \dfrac12\cdot\left(\dfrac{2x + 5 + 10 -2x}{2}\right)^2$
$\to y \leq \dfrac{1}{8}\cdot15^2 = \dfrac{225}{8}$
Dấu $=$ xảy ra $\Leftrightarrow 2x+ 5 = 10 - 2x \Leftrightarrow x = \dfrac{5}{4}$
Vậy $\max y = \dfrac{225}{8} \Leftrightarrow x = \dfrac{5}{4}$
f) $y = \dfrac{x}{x^2+2}\qquad (x >0)$
$\to y = \dfrac{1}{x + \dfrac{2}{x}}\leq \dfrac{1}{2\sqrt{x\cdot\dfrac{2}{x}}}$
$\to y \leq \dfrac{1}{2\sqrt2} = \dfrac{\sqrt2}{4}$
Dấu $=$ xảy ra $\Leftrightarrow \begin{cases}x = \dfrac{2}{x}\\x >0\end{cases} \Leftrightarrow x = \sqrt2$
Vậy $\max y = \dfrac{\sqrt2}{4}\Leftrightarrow x = \sqrt2$
