Giải thích các bước giải:
\(\begin{array}{l}
1,\\
A = {x^2} + 6x + 11\\
= \left( {{x^2} + 6x + 9} \right) + 2\\
= {\left( {x + 3} \right)^2} + 2\\
{\left( {x + 3} \right)^2} \ge 0,\,\,\,\forall x \Rightarrow {\left( {x + 3} \right)^2} + 2 \ge 2 > 0,\,\,\,\forall x\\
2,\\
B = - 2{x^2} + 8x - 17\\
= - \left( {2{x^2} - 8x + 8} \right) - 9\\
= - 2.\left( {{x^2} - 4x + 4} \right) - 9\\
= - 2.{\left( {x - 2} \right)^2} - 9\\
{\left( {x - 2} \right)^2} \ge 0,\,\,\,\,\forall x\\
\Rightarrow - 2{\left( {x - 2} \right)^2} \le 0,\,\,\,\forall x\\
\Rightarrow - 2{\left( {x - 2} \right)^2} - 9 \le - 9 < 0,\,\,\,\forall x\\
\Rightarrow B < 0,\,\,\forall x\\
3,\\
C = 4{x^2} + 12x - 5\\
= \left( {4{x^2} + 12x + 9} \right) - 14\\
= \left[ {{{\left( {2x} \right)}^2} + 2.2x.3 + {3^2}} \right] - 14\\
= {\left( {2x + 3} \right)^2} - 14 \ge - 14,\,\,\,\forall x\\
\Rightarrow {C_{\min }} = - 14 \Leftrightarrow {\left( {2x + 3} \right)^2} = 0 \Leftrightarrow x = - \frac{3}{2}\\
4,\\
D = 12x - 5 - {x^2}\\
= \left( { - {x^2} + 12x - 36} \right) + 31\\
= 31 - \left( {{x^2} - 12x + 36} \right)\\
= 31 - {\left( {x - 6} \right)^2} \le 31,\,\,\,\forall x\\
\Rightarrow {D_{\max }} = 31 \Leftrightarrow {\left( {x - 6} \right)^2} = 0 \Leftrightarrow x = 6
\end{array}\)
