Đáp án:
$\begin{array}{l}
B6)a)A = - {x^2} - x + 1\\
= - \left( {{x^2} + x} \right) + 1\\
= - \left( {{x^2} + 2.x.\dfrac{1}{2} + \dfrac{1}{4}} \right) + \dfrac{1}{4} + 1\\
= - {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{5}{4} \le \dfrac{5}{4}\\
\Leftrightarrow GTLN:A = \dfrac{5}{4}\,khi:x = - \dfrac{1}{2}\\
b)B = - {x^2} + 2x + 1\\
= - \left( {{x^2} - 2x} \right) + 1\\
= - \left( {{x^2} - 2x + 1} \right) + 1 + 1\\
= - {\left( {x - 1} \right)^2} + 2 \le 2\\
\Leftrightarrow GTLN:B = 2\,khi:x = 1\\
c)C = - {x^2} - 4{y^2} - 4xy + 12\\
= - \left( {{x^2} + 4xy + 4{y^2}} \right) + 12\\
= - {\left( {x + 2y} \right)^2} + 12 \le 12\\
\Leftrightarrow GTLN:C = 12\,khi:x = - 2y\\
d)D = - {x^2} + 3x + 1\\
= - \left( {{x^2} - 3x} \right) + 1\\
= - \left( {{x^2} - 2.x.\dfrac{3}{2} + \dfrac{9}{4}} \right) + \dfrac{9}{4} + 1\\
= - {\left( {x - \dfrac{3}{2}} \right)^2} + \dfrac{{13}}{4} \le \dfrac{{13}}{4}\\
\Leftrightarrow GTLN:D = \dfrac{{13}}{4}\,khi:x = \dfrac{3}{2}\\
B7)a)A = {x^2} + 2{y^2} - 2xy + 2y\\
= {x^2} - 2xy + {y^2} + {y^2} + 2y + 1 - 1\\
= {\left( {x - y} \right)^2} + {\left( {y + 1} \right)^2} - 1 \ge - 1\\
\Leftrightarrow GTLN:A = - 1\,khi:x = y = - 1\\
b)B = 5{x^2} + {y^2} + 4xy + 2x + 5\\
= 4{x^2} + 4xy + {y^2} + {x^2} + 2x + 1 + 4\\
= {\left( {2x + y} \right)^2} + {\left( {x + 1} \right)^2} + 4 \ge 4\\
\Leftrightarrow GTNN:B = 4\,khi:\left\{ \begin{array}{l}
y = - 2x = 2\\
x = - 1
\end{array} \right.\\
C = {x^2} + 10{y^2} + 6xy - y + 1\\
= {x^2} + 6xy + 9{y^2} + {y^2} - y + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {x + 3y} \right)^2} + {\left( {y - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
\Leftrightarrow GTNN:C = \dfrac{3}{4}\,khi:\left\{ \begin{array}{l}
x = - 3y = \dfrac{{ - 3}}{2}\\
y = \dfrac{1}{2}
\end{array} \right.
\end{array}$
