Đáp án:
$\begin{array}{l}
4)a)\\
A = - {x^2} + 6x - 5\\
= - \left( {{x^2} - 6x + 9} \right) + 4\\
= - {\left( {x - 3} \right)^2} + 4 \le 4\\
GTLN:4,khi:x = 3\\
b)B = - 4{x^2} + 3x + 1\\
= - \left( {4{x^2} - 2.2.x.\dfrac{3}{4} + \dfrac{9}{{16}}} \right) + \dfrac{{25}}{{16}}\\
= - {\left( {2x - \dfrac{3}{4}} \right)^2} + \dfrac{{25}}{{16}} \le \dfrac{{25}}{{16}}\\
GTLN:\dfrac{{25}}{{16}},khi:x = \dfrac{3}{8}\\
c)C = x\left( {2 - 9x} \right)\\
= - 9{x^2} + 2x - \dfrac{1}{9} + \dfrac{1}{9}\\
= - \left( {3x - \dfrac{1}{3}} \right) + \dfrac{1}{9} \le \dfrac{1}{9}\\
GTLN:\dfrac{1}{9},khi:x = \dfrac{1}{9}\\
5)a)\\
A = {x^2} - 2x + 4\\
= {\left( {x - 1} \right)^2} + 3 \ge 3\\
GTNN:3,khi:x = 1\\
b)B = 4{x^2} + x - 1\\
= 4{x^2} + 2.2.\dfrac{1}{4}x + \dfrac{1}{{16}} + \dfrac{{15}}{{16}}\\
= {\left( {2x + \dfrac{1}{4}} \right)^2} + \dfrac{{15}}{{16}} \ge \dfrac{{15}}{{16}}\\
GTNN:\dfrac{{15}}{{16}},khi:x = - \dfrac{1}{8}\\
c)C = x\left( {4x - 3} \right)\\
= 4{x^2} - 2.2.\dfrac{3}{4}x + \dfrac{9}{{16}} - \dfrac{9}{{16}}\\
= {\left( {2x - \dfrac{3}{4}} \right)^2} - \dfrac{9}{{16}} \ge - \dfrac{9}{{16}}\\
GTNN: - \dfrac{9}{{16}},khi:x = \dfrac{3}{8}
\end{array}$
