Đáp án:
\(\begin{array}{l}
{A_{\min }} = - 3\\
{B_{\min }} = 10\\
{C_{\min }} = - 36\\
{D_{\max }} = 21\\
{E_{\max }} = 5
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = {\left( {x - 2} \right)^2} - 3 \ge - 3\\
\Rightarrow {A_{\min }} = - 3 \Leftrightarrow x = 2\\
B = {\left( {2x + 1} \right)^2} + 10 \ge 10\\
\Rightarrow {B_{\min }} = 10 \Leftrightarrow x = - \dfrac{1}{2}\\
D = 21 - {\left( {x + 4} \right)^2} \le 21\\
\Rightarrow {D_{\max }} = 21 \Leftrightarrow x = - 4\\
E = 5 - {\left( {x - 2} \right)^2} \le 5\\
\Rightarrow {E_{\max }} = 5 \Leftrightarrow x = 2\\
C = \left( {{x^2} + 5x - 6} \right)\left( {{x^2} + 5x + 6} \right)\\
Dat\,{x^2} + 5x = t \Rightarrow C = {t^2} - 36 \ge - 36\\
\Rightarrow {C_{\min }} = - 36 \Leftrightarrow {x^2} + 5x = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 5
\end{array} \right.
\end{array}\)
