Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
Q = {x^2} - 2x + 5 = \left( {{x^2} - 2x + 1} \right) + 4 = {\left( {x - 1} \right)^2} + 4\\
{\left( {x - 1} \right)^2} \ge 0,\,\,\,\forall x \Rightarrow {\left( {x - 1} \right)^2} + 4 \ge 4,\,\,\,\forall x\\
\Rightarrow Q \ge 4,\,\,\,\forall x\\
\Rightarrow {Q_{\min }} = 4 \Leftrightarrow {\left( {x - 1} \right)^2} = 0 \Leftrightarrow x = 1\\
2,\\
M = - {x^2} + 6x + 1 = - \left( {{x^2} - 6x + 9} \right) + 10 = 10 - {\left( {x - 3} \right)^2}\\
{\left( {x - 3} \right)^2} \ge 0,\,\,\,\forall x \Rightarrow 10 - {\left( {x - 3} \right)^2} \le 10,\,\,\,\forall x\\
\Rightarrow M \le 10,\,\,\,\forall x\\
\Rightarrow {M_{\max }} = 10 \Leftrightarrow {\left( {x - 3} \right)^2} = 0 \Leftrightarrow x = 3\\
3,\\
N = 2{x^2} - 2x = 2.\left( {{x^2} - x + \dfrac{1}{4}} \right) - \dfrac{1}{2}\\
= 2.\left( {{x^2} - 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right) - \dfrac{1}{2} = 2.{\left( {x - \dfrac{1}{2}} \right)^2} - \dfrac{1}{2}\\
{\left( {x - \dfrac{1}{2}} \right)^2} \ge 0,\,\,\,\forall x \Rightarrow 2.{\left( {x - \dfrac{1}{2}} \right)^2} - \dfrac{1}{2} \ge - \dfrac{1}{2},\,\,\,\forall x\\
\Rightarrow N \ge - \dfrac{1}{2},\,\,\,\forall x\\
\Rightarrow {N_{\min }} = - \dfrac{1}{2} \Leftrightarrow {\left( {x - \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x = \dfrac{1}{2}\\
4,\\
E = 3x - 4{x^2} = - \left( {4{x^2} - 3x + \dfrac{9}{{16}}} \right) + \dfrac{9}{{16}}\\
= \dfrac{9}{{16}} - \left[ {{{\left( {2x} \right)}^2} - 2.2x.\dfrac{3}{4} + {{\left( {\dfrac{3}{4}} \right)}^2}} \right] = \dfrac{9}{{16}} - {\left( {2x - \dfrac{3}{4}} \right)^2}\\
{\left( {2x - \dfrac{3}{4}} \right)^2} \ge 0,\,\,\forall x \Rightarrow \dfrac{9}{{16}} - {\left( {2x - \dfrac{3}{4}} \right)^2} \le \dfrac{9}{{16}},\,\,\,\forall x\\
\Rightarrow E \le \dfrac{9}{{16}},\,\,\,\forall x\\
\Rightarrow {E_{\max }} = \dfrac{9}{{16}} \Leftrightarrow {\left( {2x - \dfrac{3}{4}} \right)^2} = 0 \Leftrightarrow x = \dfrac{3}{8}
\end{array}\)
