Đáp án:
$\begin{array}{l}
a)A = 5{x^2} + 12 - 2x\\
= 5.\left( {{x^2} - \dfrac{2}{5}.x} \right) + 12\\
= 5.\left( {{x^2} - 2.x.\dfrac{1}{5} + \dfrac{1}{{25}}} \right) - 5.\dfrac{1}{{25}} + 12\\
= 5.{\left( {x - \dfrac{1}{5}} \right)^2} + \dfrac{{59}}{5} \ge \dfrac{{59}}{5}\\
\Leftrightarrow GTNN:A = \dfrac{{59}}{5}\,khi:x = \dfrac{1}{5}\\
b)B = 15 - 2{x^2} + 3x\\
= - 2\left( {{x^2} - \dfrac{3}{2}.x} \right) + 15\\
= - 2.\left( {{x^2} - 2.x.\dfrac{3}{4} + \dfrac{9}{{16}}} \right) + 2.\dfrac{9}{{16}} + 15\\
= - 2.{\left( {x - \dfrac{3}{4}} \right)^2} + \dfrac{{99}}{8} \le \dfrac{{99}}{8}\\
\Leftrightarrow B \le \dfrac{{99}}{8}\\
\Leftrightarrow GTLN:B = \dfrac{{99}}{8}\,khi:x = \dfrac{3}{4}
\end{array}$
