Đáp án:
$\begin{array}{l}
1)a)A = {x^2} + x - 2\\
= {x^2} + 2.x.\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{9}{4}\\
= {\left( {x + \dfrac{1}{2}} \right)^2} - \dfrac{9}{4} \ge \dfrac{{ - 9}}{4}\\
\Leftrightarrow GTNN:A = - \dfrac{9}{4}\,khi:x = - \dfrac{1}{2}\\
b)B = {x^2} + {y^2} + x - 6y + 5\\
= {x^2} + x + {y^2} - 6y + 5\\
= {x^2} + 2.x.\dfrac{1}{2} + \dfrac{1}{4} + {y^2} - 6y + 9 - \dfrac{{17}}{4}\\
= {\left( {x + \dfrac{1}{2}} \right)^2} + {\left( {y - 3} \right)^2} - \dfrac{{17}}{4} \ge \dfrac{{ - 17}}{4}\\
\Leftrightarrow GTNN:B = - \dfrac{{17}}{4}\,khi:x = - \dfrac{1}{2};y = 3\\
2)\\
b)B = 12x - 8y - 4{x^2} - {y^2} + 1\\
= - \left( {4{x^2} - 12x} \right) - \left( {{y^2} + 8y} \right) + 1\\
= - \left( {4{x^2} - 12x + 9} \right) - \left( {{y^2} + 8y + 16} \right) + 9 + 16 + 1\\
= - {\left( {2x - 3} \right)^2} - {\left( {y + 4} \right)^2} + 26 \le 26\\
\Leftrightarrow GTLN:B = 26\,khi:\left\{ \begin{array}{l}
x = \dfrac{3}{2}\\
y = - 4
\end{array} \right.\\
c)C = 6{x^2} - {x^2} + 3\\
= 5{x^2} + 3 \ge 3\\
\Leftrightarrow GTNN:C = 3\,khi:x = 0\\
d)D = 2x - 6y - {x^2} - {y^2} - 2\\
= - \left( {{x^2} - 2x} \right) - \left( {{y^2} + 6y} \right) - 2\\
= - \left( {{x^2} - 2.x + 1} \right) - \left( {{y^2} + 6y + 9} \right) + 1 + 9 - 2\\
= - {\left( {x - 1} \right)^2} - {\left( {y + 3} \right)^2} + 8\\
\Leftrightarrow GTLN:D = 8\,khi:x = 1;y = - 3
\end{array}$
