Đáp án:
$\begin{array}{l}
a)4x - {x^2} - 5\\
= - \left( {{x^2} - 4x} \right) - 5\\
= - \left( {{x^2} - 4x + 4} \right) + 4 - 5\\
= - {\left( {x - 2} \right)^2} - 1 \le - 1\\
\Leftrightarrow GTLN = - 1\,khi:x = 2\\
b)x - {x^2}\\
= - \left( {{x^2} - x} \right)\\
= - \left( {{x^2} - 2.x.\frac{1}{2} + \frac{1}{4}} \right) + \frac{1}{4}\\
= - {\left( {x - \frac{1}{2}} \right)^2} + \frac{1}{4} \le \frac{1}{4}\\
\Leftrightarrow GTLN = \frac{1}{4}\,khi:x = \frac{1}{2}\\
c)2x - 2{x^2} - 8\\
= - 2\left( {{x^2} - x} \right) - 8\\
= - 2\left( {{x^2} - 2.x.\frac{1}{2} + \frac{1}{4}} \right) + 2.\frac{1}{4} - 8\\
= - 2{\left( {x - \frac{1}{2}} \right)^2} - \frac{{15}}{2} \le \frac{{ - 15}}{2}\\
\Leftrightarrow GTLN = - \frac{{15}}{2}\,khi:x = \frac{1}{2}\\
d)5 - 8x - {x^2}\\
= - \left( {{x^2} + 8x} \right) + 5\\
= - \left( {{x^2} + 8x + 16} \right) + 16 + 5\\
= - {\left( {x + 4} \right)^2} + 21 \le 21\\
\Leftrightarrow GTLN = 21\,khi:x = - 4\\
e)4x - {x^2} + 1\\
= - \left( {{x^2} - 4x} \right) + 1\\
= - \left( {{x^2} - 2.x.2 + 4} \right) + 4 + 1\\
= - {\left( {x - 2} \right)^2} + 5 \le 5\\
\Leftrightarrow GTLN = 5\,khi:x = 2\\
f)6x - 3{x^2} + 1\\
= - 3\left( {{x^2} - 2x} \right) + 1\\
= - 3\left( {{x^2} - 2x + 1} \right) + 3 + 1\\
= - 3{\left( {x - 1} \right)^2} + 4 \le 4\\
\Leftrightarrow GTLN = 4\,khi:x = 1
\end{array}$
