Đáp án:
$\begin{array}{l}
VD2:\\
a)y = 4x\left( {8 - 5x} \right)\\
= - 20{x^2} + 32x\\
= - \left( {20{x^2} - 32x} \right)\\
= - \left( {{{\left( {2\sqrt 5 x} \right)}^2} - 2.2\sqrt 5 x.\frac{8}{{\sqrt 5 }} + \frac{{64}}{5}} \right) + \frac{{64}}{5}\\
= - {\left( {2\sqrt 5 x - \frac{8}{{\sqrt 5 }}} \right)^2} + \frac{{64}}{5} \le \frac{{64}}{5}\forall 0 \le x \le \frac{8}{5}\\
\Rightarrow GTLN:y = \frac{{64}}{5} \Leftrightarrow x = \frac{4}{5}\\
b)y = \left( {2x - 1} \right).\left( {3 - x} \right)\\
\Rightarrow y = \frac{1}{2}.\left( {2x - 1} \right).\left( {6 - 2x} \right) \le \frac{1}{2}.\frac{{{{\left( {2x - 1 + 6 - 2x} \right)}^2}}}{4} = \frac{{25}}{8}\\
\Rightarrow GTLN:y = \frac{{25}}{8} \Leftrightarrow 2x - 1 = 6 - 2x \Rightarrow x = \frac{7}{4}\left( {tmdk} \right)\\
VD3:\\
{x^2} + \frac{2}{{{x^3}}}\\
= \frac{1}{3}{x^2} + \frac{1}{3}{x^2} + \frac{1}{3}{x^2} + \frac{1}{{{x^3}}} + \frac{1}{{{x^3}}}\\
\ge 6.\sqrt[6]{{\frac{1}{3}{x^2}.\frac{1}{3}{x^2}.\frac{1}{3}{x^2}.\frac{1}{{{x^3}}}.\frac{1}{{{x^3}}}}} = 6.\frac{1}{{\sqrt 3 }} = 2\sqrt 3 \\
\Rightarrow GTNN:y = 2\sqrt 3 \Leftrightarrow \frac{1}{3}{x^2} = \frac{1}{{{x^3}}} \Rightarrow {x^5} = 3 \Rightarrow x = \sqrt[5]{3}
\end{array}$
