Giải thích các bước giải:

$\begin{array}{l}
B7:\\
a)A = {x^2} - 2x + 5 = {\left( {x - 1} \right)^2} + 4 \ge 4,\forall x\\
 \Rightarrow MinA = 4 \Leftrightarrow x = 1\\
c)C = \left( {x - 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 6} \right)\\
 = \left( {x - 1} \right)\left( {x + 6} \right)\left( {x + 2} \right)\left( {x + 3} \right)\\
 = \left( {{x^2} + 5x - 6} \right)\left( {{x^2} + 5x + 6} \right)\\
 = {\left( {{x^2} + 5x} \right)^2} - {6^2}\\
 = {\left( {{x^2} + 5x} \right)^2} - 36 \ge  - 36,\forall x\\
 \Rightarrow MinC =  - 36 \Leftrightarrow {x^2} + 5x = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x =  - 5
\end{array} \right.\\
d)D = {x^2} + 5{y^2} - 2xy + 4y + 3\\
 = \left( {{x^2} - 2xy + {y^2}} \right) + \left( {4{y^2} + 4y + 1} \right) + 2\\
 = {\left( {x - y} \right)^2} + {\left( {2y + 1} \right)^2} + 2 \ge 2,\forall x,y\\
 \Rightarrow MinD = 2 \Leftrightarrow \left\{ \begin{array}{l}
x - y = 0\\
2y + 1 = 0
\end{array} \right. \Leftrightarrow x = y = \frac{{ - 1}}{2}\\
B8:\\
a)A =  - {x^2} - 4x - 2\\
 =  - \left( {{x^2} + 4x + 4} \right) + 2\\
 =  - {\left( {x + 2} \right)^2} + 2 \le 2,\forall x\\
 \Rightarrow MaxA = 2 \Leftrightarrow x + 2 = 0 \Leftrightarrow x =  - 2\\
d)D =  - 8{x^2} + 4xy - {y^2} + 3\\
 = \frac{{ - 1}}{2}\left( {16{x^2} - 8xy + {y^2}} \right) - \frac{1}{2}{y^2} + 3\\
 = \frac{{ - 1}}{2}{\left( {4x - y} \right)^2} - \frac{1}{2}{y^2} + 3 \ge 3,\forall x,y\\
 \Rightarrow MaxD = 3 \Leftrightarrow \left\{ \begin{array}{l}
4x - y = 0\\
y = 0
\end{array} \right. \Leftrightarrow x = y = 0\\
B9:\\
a)A = 25{x^2} - 20x + 7\\
 = {\left( {5x} \right)^2} - 2.5x.2 + 4 + 3\\
 = {\left( {5x - 2} \right)^2} + 3 \ge 3,\forall x\\
 \Rightarrow A > 0,\forall x\\
c)E = {x^2} - 2x + {y^2} + 4y + 6\\
 = \left( {{x^2} - 2x + 1} \right) + \left( {{y^2} + 4y + 4} \right) + 1\\
 = {\left( {x - 1} \right)^2} + {\left( {y + 2} \right)^2} + 1 \ge 1,\forall x\\
 \Rightarrow E > 0,\forall x\\
B10:\\
n \in N\\
A = n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right) + 1\\
 = n\left( {n + 3} \right)\left( {n + 1} \right)\left( {n + 2} \right) + 1\\
 = \left( {{n^2} + 3n} \right)\left( {{n^2} + 3n + 2} \right) + 1\\
 = {\left( {{n^2} + 3n} \right)^2} + 2\left( {{n^2} + 3n} \right) + 1\\
 = {\left( {{n^2} + 3n + 1} \right)^2}
\end{array}$

Suy ra A là số chính phương.