Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
D = 4 - {x^2} + 2x\\
= 5 + \left( { - {x^2} + 2x - 1} \right)\\
= 5 - \left( {{x^2} - 2x + 1} \right)\\
= 5 - {\left( {x - 1} \right)^2}\\
{\left( {x - 1} \right)^2} \ge 0,\,\,\,\forall x \Rightarrow 5 - {\left( {x - 1} \right)^2} \le 5,\,\,\,\forall x\\
\Rightarrow {D_{\max }} = 5 \Leftrightarrow {\left( {x - 1} \right)^2} = 0 \Leftrightarrow x = 1\\
b,\\
E = \left( {x + 4} \right)\left( {2 - x} \right) - 10\\
= \left( {2x - {x^2} + 8 - 4x} \right) - 10\\
= - {x^2} - 2x - 2\\
= \left( { - {x^2} - 2x - 1} \right) - 1\\
= - 1 - \left( {{x^2} + 2x + 1} \right)\\
= - 1 - {\left( {x + 1} \right)^2}\\
{\left( {x + 1} \right)^2} \ge 0,\,\,\forall x \Rightarrow - 1 - {\left( {x + 1} \right)^2} \le - 1,\,\,\,\forall x\\
\Rightarrow {E_{\max }} = - 1 \Leftrightarrow {\left( {x + 1} \right)^2} = 0 \Leftrightarrow x = - 1\\
c,\\
F = - {x^2} + 6x - 15\\
= \left( { - {x^2} + 6x - 9} \right) - 6\\
= - 6 - \left( {{x^2} - 6x + 9} \right)\\
= - 6 - {\left( {x - 3} \right)^2}\\
{\left( {x - 3} \right)^2} \ge 0,\,\,\forall x \Rightarrow - 6 - {\left( {x - 3} \right)^2} \le - 6,\,\,\,\forall x\\
\Rightarrow {F_{\max }} = - 6 \Leftrightarrow {\left( {x - 3} \right)^2} = 0 \Leftrightarrow x = 3
\end{array}\)
