Đáp án:
$\begin{array}{l}
a)A = {\left( {x - 2} \right)^3} - {x^2}\left( {x - 4} \right) + 8\\
= {x^3} + 3.{x^2}.\left( { - 2} \right) + 3.x.{\left( { - 2} \right)^2} + {\left( { - 2} \right)^3}\\
- {x^3} + 4{x^2} + 8\\
= - 6{x^2} + 12x - 8 + 4{x^2} + 8\\
= - 2{x^2} + 12x\\
B = \left( {{x^2} - 6x + 9} \right):\left( {x - 3} \right) - x\left( {x + 7} \right) - 9\\
= {\left( {x - 3} \right)^2}:\left( {x - 3} \right) - {x^2} - 7x - 9\\
= x - 3 - {x^2} - 7x - 9\\
= - {x^2} - 6x - 12\\
b)Khi:x = - 1\\
\Rightarrow A = - 2{x^2} + 12x\\
= - 2.{\left( { - 1} \right)^2} + 12.\left( { - 1} \right)\\
= - 2.1 - 12\\
= - 14\\
c)C = A + B\\
= - 2{x^2} + 12x - {x^2} - 6x - 12\\
= - 3{x^2} + 6x - 12\\
= - 3.\left( {{x^2} - 2x} \right) - 12\\
= - 3.\left( {{x^2} - 2x + 1} \right) + 3 - 12\\
= - 3{\left( {x - 1} \right)^2} - 9\\
Do:{\left( {x - 1} \right)^2} \ge 0\\
\Rightarrow - 3{\left( {x - 1} \right)^2} \le 0\\
\Rightarrow - 3{\left( {x - 1} \right)^2} - 9 \le - 9 < 0\\
\Rightarrow C < 0\\
Vậy\,C < 0\,\forall x \ne 3
\end{array}$
