Đáp án:

\(\begin{array}{l}
3,\\
a,\\
{M_{\min }} = 1 \Leftrightarrow x = 2\\
b,\\
{N_{\min }} =  - \dfrac{{13}}{4} \Leftrightarrow y = \dfrac{1}{2}\\
4,\\
a,\\
{A_{\max }} = 6 \Leftrightarrow x = 2\\
b,\\
{B_{\max }} = \dfrac{{25}}{4} \Leftrightarrow x = \dfrac{1}{2}
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
3,\\
a,\\
M = {x^2} - 4x + 5 = \left( {{x^2} - 4x + 4} \right) + 1\\
 = \left( {{x^2} - 2.x.2 + {2^2}} \right) + 1 = {\left( {x - 2} \right)^2} + 1\\
{\left( {x - 2} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow M = {\left( {x - 2} \right)^2} + 1 \ge 1,\,\,\,\forall x\\
 \Rightarrow {M_{\min }} = 1 \Leftrightarrow {\left( {x - 2} \right)^2} = 0 \Leftrightarrow x - 2 = 0 \Leftrightarrow x = 2\\
 \Rightarrow {M_{\min }} = 1 \Leftrightarrow x = 2\\
b,\\
N = {y^2} - y - 3 = \left( {{y^2} - y + \dfrac{1}{4}} \right) - \dfrac{{13}}{4}\\
 = \left( {{y^2} - 2.y.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right) - \dfrac{{13}}{4}\\
 = {\left( {y - \dfrac{1}{2}} \right)^2} - \dfrac{{13}}{4}\\
{\left( {y - \dfrac{1}{2}} \right)^2} \ge 0,\,\,\,\forall y\\
 \Rightarrow N = {\left( {y - \dfrac{1}{2}} \right)^2} - \dfrac{{13}}{4} \ge  - \dfrac{{13}}{4},\,\,\,\forall y\\
 \Rightarrow {N_{\min }} =  - \dfrac{{13}}{4} \Leftrightarrow {\left( {y - \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow y - \dfrac{1}{2} = 0 \Leftrightarrow y = \dfrac{1}{2}\\
 \Rightarrow {N_{\min }} =  - \dfrac{{13}}{4} \Leftrightarrow y = \dfrac{1}{2}\\
4,\\
a,\\
A =  - {x^2} + 4x + 2 = 6 + \left( { - {x^2} + 4x - 4} \right)\\
 = 6 - \left( {{x^2} - 4x + 4} \right) = 6 - \left( {{x^2} - 2.x.2 + {2^2}} \right)\\
 = 6 - {\left( {x - 2} \right)^2}\\
{\left( {x - 2} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow A = 6 - {\left( {x - 2} \right)^2} \le 6,\,\,\,\forall x\\
 \Rightarrow {A_{\max }} = 6 \Leftrightarrow {\left( {x - 2} \right)^2} = 0 \Leftrightarrow x - 2 = 0 \Leftrightarrow x = 2\\
b,\\
B = x - {x^2} + 6 = \dfrac{{25}}{4} + \left( { - {x^2} + x - \dfrac{1}{4}} \right)\\
 = \dfrac{{25}}{4} - \left( {{x^2} - x + \dfrac{1}{4}} \right)\\
 = \dfrac{{25}}{4} - \left( {{x^2} - 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right)\\
 = \dfrac{{25}}{4} - {\left( {x - \dfrac{1}{2}} \right)^2}\\
{\left( {x - \dfrac{1}{2}} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow B = \dfrac{{25}}{4} - {\left( {x - \dfrac{1}{2}} \right)^2} \le \dfrac{{25}}{4},\,\,\,\forall x\\
 \Rightarrow {B_{\max }} = \dfrac{{25}}{4} \Leftrightarrow {\left( {x - \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x - \dfrac{1}{2} = 0 \Leftrightarrow x = \dfrac{1}{2}\\
 \Rightarrow {B_{\max }} = \dfrac{{25}}{4} \Leftrightarrow x = \dfrac{1}{2}
\end{array}\)