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