Đáp án:
$\begin{array}{l}
a/\\
{A_{\min }} = - 7\\
{B_{\min }} = \frac{{31}}{{16}}\\
b/\\
{A_{\max }} = \frac{{37}}{4}\\
{C_{\max }} = 2018
\end{array}$
Giải thích các bước giải:a/
$\begin{array}{l}
A = {x^2} - 6x + 2\\
= ({x^2} - 3.2x + {3^2}) - {3^2} + 2\\
= {(x - 3)^2} - 7 \ge - 7
\end{array}$
Dấu "=" xảy ra khi x=3
$\begin{array}{l}
B = 4{x^2} - x + 2\\
= {(2x)^2} - 2.2x.\frac{1}{4} + {\left( {\frac{1}{4}} \right)^2} - {\left( {\frac{1}{4}} \right)^2} + 2\\
= {\left( {2x - \frac{1}{4}} \right)^2} + \frac{{31}}{{16}} \ge \frac{{31}}{{16}}
\end{array}$
Dấu "="xảy ra khi
$x = \frac{1}{8}$
b/
$\begin{array}{l}
A = - {x^2} + 5x + 3\\
= - \left[ {{x^2} - 2.x.\frac{5}{2} + {{\left( {\frac{5}{2}} \right)}^2}} \right] + {\left( {\frac{5}{2}} \right)^2} + 3\\
= \frac{{37}}{4} - {\left( {x - \frac{5}{2}} \right)^2} \le \frac{{37}}{4}
\end{array}$
dấu "=" xảy ra khi
$x = \frac{5}{2}$
+
$\begin{array}{l}
C = \frac{{2018}}{{{x^2} - 22x + 122}}\\
{x^2} - 22x + 122 = {x^2} - 2.x.11 + 121 + 1 = {(x - 11)^2} + 1 \ge 1\\
= > C = \frac{{2018}}{{{{(x - 11)}^2} + 1}} \le \frac{{2018}}{1} = 2018
\end{array}$
Dấu"=" xảy ra khi x=11
