Đáp án:

\(\begin{array}{l}
a,\\
{A_{\min }} = 2 \Leftrightarrow \left\{ \begin{array}{l}
x =  - 1\\
y = 2
\end{array} \right.\\
b,\\
{B_{\min }} = 2 \Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y =  - \dfrac{1}{2}\\
z = 4
\end{array} \right.\\
c,\\
{C_{\min }} = 1 \Leftrightarrow x = 1\\
d,\\
{D_{\min }} = 4 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x =  - 5
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
A = {x^2} + 2x + {y^2} - 4y + 7\\
 = \left( {{x^2} + 2x + 1} \right) + \left( {{y^2} - 4y + 4} \right) + 2\\
 = \left( {{x^2} + 2.x.1 + {1^2}} \right) + \left( {{y^2} - 2.y.2 + {2^2}} \right) + 2\\
 = {\left( {x + 1} \right)^2} + {\left( {y - 2} \right)^2} + 2\\
{\left( {x + 1} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {y - 2} \right)^2} \ge 0,\,\,\,\forall y\\
 \Rightarrow {\left( {x + 1} \right)^2} + {\left( {y - 2} \right)^2} \ge 0,\,\,\,\forall x,y\\
 \Rightarrow A = {\left( {x + 1} \right)^2} + {\left( {y - 2} \right)^2} + 2 \ge 2,\,\,\,\forall x,y\\
 \Rightarrow {A_{\min }} = 2 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + 1} \right)^2} = 0\\
{\left( {y - 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x =  - 1\\
y = 2
\end{array} \right.\\
b,\\
B = {x^2} + 4{y^2} + {z^2} - 4x + 4y - 8z + 23\\
 = \left( {{x^2} - 4x + 4} \right) + \left( {4{y^2} + 4y + 1} \right) + \left( {{z^2} - 8z + 16} \right) + 2\\
 = \left( {{x^2} - 2.x.2 + {2^2}} \right) + \left[ {{{\left( {2y} \right)}^2} + 2.2y.1 + {1^2}} \right] + \left( {{z^2} - 2.z.4 + {4^2}} \right) + 2\\
 = {\left( {x - 2} \right)^2} + {\left( {2y + 1} \right)^2} + {\left( {z - 4} \right)^2} + 2\\
{\left( {x - 2} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {2y + 1} \right)^2} \ge 0,\,\,\,\forall y\\
{\left( {z - 4} \right)^2} \ge 0,\,\,\,\forall z\\
 \Rightarrow {\left( {x - 2} \right)^2} + {\left( {2y + 1} \right)^2} + {\left( {z - 4} \right)^2} \ge 0,\,\,\,\forall x,y,z\\
 \Rightarrow B = {\left( {x - 2} \right)^2} + {\left( {2y + 1} \right)^2} + {\left( {z - 4} \right)^2} + 2 \ge 2,\,\,\,\forall x,y,z\\
 \Rightarrow {B_{\min }} = 2 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - 2} \right)^2} = 0\\
{\left( {2y + 1} \right)^2} = 0\\
{\left( {z - 4} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y =  - \dfrac{1}{2}\\
z = 4
\end{array} \right.\\
c,\\
C = \left( {{x^2} - 2x} \right)\left( {{x^2} - 2x + 2} \right) + 2\\
 = \left[ {\left( {{x^2} - 2x + 1} \right) - 1} \right].\left[ {\left( {{x^2} - 2x + 1} \right) + 1} \right] + 2\\
 = {\left( {{x^2} - 2x + 1} \right)^2} - {1^2} + 2\\
 = {\left( {{x^2} - 2x.1 + {1^2}} \right)^2} + 1\\
 = {\left[ {{{\left( {x - 1} \right)}^2}} \right]^2} + 1\\
 = {\left( {x - 1} \right)^4} + 1\\
{\left( {x - 1} \right)^4} \ge 0,\,\,\,\forall x\\
 \Rightarrow C = {\left( {x - 1} \right)^4} + 1 \ge 1,\,\,\,\forall x\\
 \Rightarrow {C_{\min }} = 1 \Leftrightarrow {\left( {x - 1} \right)^4} = 0 \Leftrightarrow x - 1 = 0 \Leftrightarrow x = 1\\
d,\\
D = \left( {x - 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 6} \right) + 40\\
 = \left[ {\left( {x - 1} \right)\left( {x + 6} \right)} \right].\left[ {\left( {x + 2} \right)\left( {x + 3} \right)} \right] + 40\\
 = \left( {{x^2} + 6x - x - 6} \right).\left( {{x^2} + 3x + 2x + 6} \right) + 40\\
 = \left( {{x^2} + 5x - 6} \right).\left( {{x^2} + 5x + 6} \right) + 40\\
 = \left[ {\left( {{x^2} + 5x} \right) - 6} \right].\left[ {\left( {{x^2} + 5x} \right) + 6} \right] + 40\\
 = {\left( {{x^2} + 5x} \right)^2} - {6^2} + 40\\
 = {\left( {{x^2} + 5x} \right)^2} + 4\\
{\left( {{x^2} + 5x} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow D = {\left( {{x^2} + 5x} \right)^2} + 4 \ge 4,\,\,\,\forall x\\
 \Rightarrow {D_{\min }} = 4 \Leftrightarrow {\left( {{x^2} + 5x} \right)^2} = 0 \Leftrightarrow {x^2} + 5x = 0 \Leftrightarrow x\left( {x + 5} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x =  - 5
\end{array} \right.
\end{array}\)