Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = {x^2} + 8x = \left( {{x^2} + 8x + 16} \right) - 16\\
= \left( {{x^2} + 2.x.4 + {4^2}} \right) - 16 = {\left( {x + 4} \right)^2} - 16\\
{\left( {x + 4} \right)^2} \ge 0,\,\,\,\forall x \Rightarrow {\left( {x + 4} \right)^2} - 16 \ge - 16,\,\,\,\forall x\\
\Rightarrow {A_{\min }} = - 16 \Leftrightarrow {\left( {x + 4} \right)^2} = 0 \Leftrightarrow x = - 4\\
b,\\
B = - 2{x^2} + 8x - 15 = - \left( {2{x^2} - 8x + 8} \right) - 7\\
= - 7 - 2.\left( {{x^2} - 4x + 4} \right) = - 7 - 2.{\left( {x - 2} \right)^2}\\
{\left( {x - 2} \right)^2} \ge 0,\,\,\forall x \Rightarrow - 7 - 2{\left( {x - 2} \right)^2} \le - 7,\,\,\,\forall x\\
\Rightarrow {B_{\max }} = - 7 \Leftrightarrow {\left( {x - 2} \right)^2} = 0 \Leftrightarrow x = 2\\
c,\\
C = {x^2} - 6x + {y^2} - 2y + 12\\
= \left( {{x^2} - 6x + 9} \right) + \left( {{y^2} - 2y + 1} \right) + 2\\
= {\left( {x - 3} \right)^2} + {\left( {y - 1} \right)^2} + 2\\
{\left( {x - 3} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {y - 1} \right)^2} \ge 0,\,\,\forall y\\
\Rightarrow {\left( {x - 3} \right)^2} + {\left( {y - 1} \right)^2} + 2 \ge 2,\,\,\forall x,y\\
\Rightarrow {C_{\min }} = 2 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - 3} \right)^2} = 0\\
{\left( {y - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = 1
\end{array} \right.\\
d,\\
D = \left( {{x^2} - 4x - 5} \right)\left( {{x^2} - 4x - 19} \right) + 49\\
= \left[ {\left( {{x^2} - 4x - 12} \right) + 7} \right].\left[ {\left( {{x^2} - 4x - 12} \right) - 7} \right] + 49\\
= {\left( {{x^2} - 4x - 12} \right)^2} - {7^2} + 49\\
= {\left( {{x^2} - 4x - 12} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow {D_{\min }} = 0 \Leftrightarrow {\left( {{x^2} - 4x - 12} \right)^2} = 0 \Leftrightarrow {x^2} - 4x - 12 = 0 \Leftrightarrow \left( {x - 6} \right)\left( {x + 2} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 6\\
x = - 2
\end{array} \right.\\
e,\\
E = x - {x^2} = - \left( {{x^2} - x + \dfrac{1}{4}} \right) + \dfrac{1}{4}\\
= \dfrac{1}{4} - \left( {{x^2} - 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right)\\
= \dfrac{1}{4} - {\left( {x - \dfrac{1}{2}} \right)^2} \le \dfrac{1}{4},\,\,\,\forall x\\
\Rightarrow {E_{\max }} = \dfrac{1}{4} \Leftrightarrow {\left( {x - \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x = \dfrac{1}{2}
\end{array}\)
