Đáp án:
$\begin{array}{l}
C = \left( {x + 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 6} \right)\\
= \left( {x + 1} \right)\left( {x - 6} \right)\left( {x - 2} \right)\left( {x - 3} \right)\\
= \left( {{x^2} - 6x + x - 6} \right)\left( {{x^2} - 2x - 3x + 6} \right)\\
= \left( {{x^2} - 5x - 6} \right)\left( {{x^2} - 5x + 6} \right)\\
\text{Đặt}:{x^2} - 5x - 6 = a\\
\Rightarrow C = a\left( {a + 12} \right)\\
= {a^2} + 12a + 36 - 36\\
= {\left( {a + 6} \right)^2} - 36\\
\Rightarrow C \ge - 36\\
\Rightarrow GTNN:C = - 36\\
\text{Dấu = xảy ra} :a + 6 = 0\\
\Rightarrow {x^2} - 5x + 6 = 0\\
\Rightarrow \left( {x - 2} \right)\left( {x - 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 3
\end{array} \right.\\
D = {x^2} - 4xy + 5{y^2} + 10x - 22y + 28\\
= {x^2} + 4{y^2} + 25 + 10x - 20y - 4xy\\
+ {y^2} - 2y + 1 + 2\\
= {\left( {x - 2y + 5} \right)^2} + {\left( {y - 1} \right)^2} + 2\\
Do:{\left( {x - 2y + 5} \right)^2};{\left( {y - 1} \right)^2} \ge 0\\
\Rightarrow {\left( {x - 2y + 5} \right)^2} + {\left( {y - 1} \right)^2} \ge 0\\
\Rightarrow {\left( {x - 2y + 5} \right)^2} + {\left( {y - 1} \right)^2} + 2 \ge 2\\
\Rightarrow GTNN:D = 2\\
Khi:\left\{ \begin{array}{l}
x - 2y + 5 = 0\\
y - 1 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 2y - 5\\
y = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = - 3\\
y = 1
\end{array} \right.
\end{array}$
