Đáp án:
$\begin{array}{l}
11)\\
A = \left( {3x + 1} \right)\left( {2 - x} \right) - 20\\
= 6x - 3{x^2} + 2 - x - 20\\
= - 3{x^2} + 5x - 18\\
= - 3.\left( {{x^2} - \dfrac{5}{3}x} \right) - 18\\
= - 3.\left( {{x^2} - 2.x.\dfrac{5}{6} + \dfrac{{25}}{{36}}} \right) + 3.\dfrac{{25}}{{36}} - 18\\
= - 3.{\left( {x - \dfrac{5}{6}} \right)^2} - \dfrac{{191}}{{12}} \le - \dfrac{{191}}{{12}}\\
\Leftrightarrow GTLN:A = - \dfrac{{191}}{{12}}\,khi:x = \dfrac{5}{6}\\
12)\\
A = - {\left( {2x + 1} \right)^2} + x\left( {x - 1} \right) - 1\\
= - 4{x^2} - 4x - 1 + {x^2} - x - 1\\
= - 3{x^2} - 5x - 2\\
= - 3.\left( {{x^2} + \dfrac{5}{3}x} \right) - 2\\
= - 3.\left( {{x^2} + 2.x.\dfrac{5}{6} + \dfrac{{25}}{{36}}} \right) + 3.\dfrac{{25}}{{36}} - 2\\
= - 3.{\left( {x + \dfrac{5}{6}} \right)^2} + \dfrac{1}{{12}} \le \dfrac{1}{{12}}\\
\Leftrightarrow GTLN:A = \dfrac{1}{{12}}\,khi:x = - \dfrac{5}{6}\\
13)\\
A = - 2{x^2} + 4xy - 4{y^2} - 2x - 1\\
= - {x^2} + 4xy - 4{y^2} - {x^2} - 2x - 1\\
= - {\left( {x - 2y} \right)^2} - {\left( {x + 1} \right)^2} \le 0\\
\Leftrightarrow GTLN:A = 0\,khi:\left\{ \begin{array}{l}
x = 2y\\
x = - 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
y = - \dfrac{1}{2}\\
x = - 1
\end{array} \right.
\end{array}$
