Đáp án:
$\begin{array}{l}
4)A = - 2{x^2} - 4x + 5\\
= - 2\left( {{x^2} + 2x + 1} \right) + 2 + 5\\
= - 2{\left( {x + 1} \right)^2} + 7 \le 7\\
\Leftrightarrow GTLN:A = 7\,khi:x = - 1\\
5)\\
A = 4x - {x^2}\\
= - \left( {{x^2} - 4x + 4} \right) + 4\\
= - {\left( {x - 2} \right)^2} + 4 \le 4\\
\Leftrightarrow GTLN:A = 4\,khi:x = 2\\
6)B = {\left( {x + 2} \right)^2} - 5\left( {1 + {x^2} - 2x} \right)\\
= {x^2} + 4x + 4 - 5 - 5{x^2} + 10x\\
= - 4{x^2} + 14x - 1\\
= - \left( {4{x^2} - 14x} \right) - 1\\
= - \left( {4{x^2} - 2.2x.\dfrac{7}{2} + \dfrac{{49}}{4}} \right) + \dfrac{{49}}{4} - 1\\
= - {\left( {2x - \dfrac{7}{2}} \right)^2} + \dfrac{{45}}{4} \le \dfrac{{45}}{4}\\
\Leftrightarrow GTLN:B = \dfrac{{45}}{4}\,khi:x = \dfrac{7}{4}
\end{array}$
