Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
C = {\left( {x - 1} \right)^2}\left( {{x^2} - 2x - 1} \right)\\
= \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 2x - 1} \right)\\
= \left[ {\left( {{x^2} - 2x} \right) + 1} \right].\left[ {\left( {{x^2} - 2x} \right) - 1} \right]\\
= {\left( {{x^2} - 2x} \right)^2} - 1\\
\ge - 1,\,\,\,\,\forall x\\
\Rightarrow {C_{\min }} = - 1 \Leftrightarrow {\left( {{x^2} - 2x} \right)^2} = 0 \Leftrightarrow {x^2} - 2x = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.\\
d,\\
D = \left( {x - 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 6} \right)\\
= \left[ {\left( {x - 1} \right)\left( {x + 6} \right)} \right].\left[ {\left( {x + 2} \right)\left( {x + 3} \right)} \right]\\
= \left( {{x^2} + 5x - 6} \right).\left( {{x^2} + 5x + 6} \right)\\
= \left[ {\left( {{x^2} + 5x} \right) - 6} \right].\left[ {\left( {{x^2} + 5x} \right) + 6} \right]\\
= {\left( {{x^2} + 5x} \right)^2} - 36 \ge - 36,\,\,\,\,\forall x\\
\Rightarrow {D_{\min }} = - 36 \Leftrightarrow {\left( {{x^2} + 5x} \right)^2} = 0 \Leftrightarrow {x^2} + 5x = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 5
\end{array} \right.
\end{array}\)
