Đáp án:
$\begin{array}{l}
a)A = 4 - 2{x^2}\\
= - 2{x^2} + 4 \le 4\\
\Leftrightarrow GTLN:A = 4\,khi:x = 0\\
b)B = - {x^2} + 10x - 5\\
= - \left( {{x^2} - 10x + 25} \right) + 20\\
= - {\left( {x - 5} \right)^2} + 20 \le 20\\
\Leftrightarrow GTLN:B = 20\,khi:x = 5\\
c)C = - 3{x^2} + 2x - 5\\
= - 3\left( {{x^2} - \dfrac{2}{3}x + \dfrac{1}{9}} \right) + \dfrac{1}{3} - 5\\
= - 3{\left( {x - \dfrac{1}{3}} \right)^2} - \dfrac{{14}}{3} \le - \dfrac{{14}}{3}\\
\Leftrightarrow GTLN:C = - \dfrac{{14}}{3}\,khi:x = \dfrac{1}{3}\\
d)D = - 9{x^2} + 24x - 18\\
= - \left( {9{x^2} - 24x + 16} \right) - 2\\
= - {\left( {3x - 4} \right)^2} - 2 \le - 2\\
\Leftrightarrow GTLN:d = - 2\,\,khi:x = \dfrac{4}{3}\\
f)C = {x^2} - 4x + 7\\
= {x^2} - 4x + 4 + 3\\
= {\left( {x - 2} \right)^2} + 3 \ge 3\\
\Leftrightarrow GTNN:C = 3\,khi:x = 2\\
g)D = 2{x^2} + 3x + 4\\
= 2\left( {{x^2} + \dfrac{3}{2}x + \dfrac{9}{{16}}} \right) - \dfrac{9}{8} + 4\\
= 2{\left( {x + \dfrac{3}{4}} \right)^2} + \dfrac{{23}}{8} \ge \dfrac{{23}}{8}\\
\Leftrightarrow GTNN:D = \dfrac{{23}}{8}\,khi:x = \dfrac{{ - 3}}{4}\\
h)E = {x^2} - x + 1\\
= {x^2} - 2.\dfrac{1}{2}x + \dfrac{1}{4} - \dfrac{1}{4} + 1\\
= {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
\Leftrightarrow GTNN:E = \dfrac{3}{4}\,khi:x = \dfrac{1}{2}
\end{array}$
