Đáp án:
$\begin{array}{l}
B4)\\
1)A = {x^2} - 6x + 10\\
= {x^2} - 2.3.x + 9 + 1\\
= {\left( {x - 3} \right)^2} + 1 \ge 1\\
\Leftrightarrow GTNN:A = 1\,khi:x = 3\\
2)B = {x^2} + x + 2\\
= {x^2} + 2.x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{7}{4}\\
= {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{7}{4} \ge \dfrac{7}{4}\\
\Leftrightarrow GTNN:B = \dfrac{7}{4}\,khi:x = - \dfrac{1}{2}\\
B5)1)N = - {x^2} - 6x - 10\\
= - \left( {{x^2} + 6x + 9} \right) - 1\\
= - {\left( {x + 3} \right)^2} - 1 \le - 1\\
\Leftrightarrow GTLN:N = - 1\,khi:x = - 3\\
2)\\
M = - {x^2} + 3x + 1\\
= - \left( {{x^2} - 3x} \right) + 1\\
= - \left( {{x^2} - 2.x.\dfrac{3}{2} + \dfrac{9}{4}} \right) + \dfrac{9}{4} + 1\\
= - {\left( {x - \dfrac{3}{2}} \right)^2} + \dfrac{{13}}{4} \le \dfrac{{13}}{4}\\
\Leftrightarrow GTLN:M = \dfrac{{13}}{4}\,khi:x = \dfrac{3}{2}
\end{array}$
