Đáp án:
$\begin{array}{l}
a)y = 1 + 8x - 2{x^2}\\
\Rightarrow y' = - 4x + 8 = 0\\
\Rightarrow x = 2\\
\Rightarrow GTLN:y\left( 2 \right) = 1 + 8.2 - {2.2^2} = 1\\
b)y = 4{x^3} - 3{x^4}\\
\Rightarrow y' = 12{x^2} - 12{x^3} = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
CT:x = 0;y = 0\\
CD:x = 1;y = 1
\end{array} \right.\\
c)y = \dfrac{{{{\left( {x + 2} \right)}^2}}}{x}\left( {x > 0} \right)\\
y = \dfrac{{{x^2} + 4x + 4}}{x} = x + 4 + \dfrac{4}{x}\\
Theo\,Co - si:x + \dfrac{4}{x} \ge 2\sqrt 4 = 4\\
\Rightarrow x + 4 + \dfrac{4}{x} \ge 4 + 4 = 8\\
\Rightarrow GTNN:y = 8\,khi:x = 2\\
d)y = \dfrac{{{x^2} + 2x + 3}}{{x + 2}}\left( {x > - 2} \right)\\
= x + \dfrac{3}{{x + 2}}\\
= x + 2 + \dfrac{3}{{x + 2}} - 2\\
Theo\,Co - si:\left( {x + 2} \right) + \dfrac{3}{{x + 2}} \ge 2\sqrt 3 \\
\Rightarrow x + 2 + \dfrac{3}{{x + 2}} - 2 \ge 2\sqrt 3 - 2\\
\Rightarrow y \ge 2\sqrt 3 - 2\\
\Rightarrow GTNN:y = 2\sqrt 3 - 2\\
Khi:{\left( {x + 2} \right)^2} = 3\\
\Rightarrow x = \sqrt 3 - 2
\end{array}$
