Đáp án:
\(\begin{array}{l}
a)\,\,\,A = - 2{x^2} + 12x.\\
B = - {x^2} - 6x - 12.\\
b)\,\,\,Khi\,\,\,x = - 1\,\,thi\,\,\,A = - 14.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\,\,\,A = {\left( {x - 2} \right)^3} - {x^2}\left( {x - 4} \right) + 8\\
= {x^3} - 6{x^2} + 12x - 8 - {x^3} + 4{x^2} + 8\\
= - 2{x^2} + 12x.\\
B = \left( {{x^2} - 6x + 9} \right):\left( {x - 3} \right) - x\left( {x + 7} \right) - 9\,\,\,\left( {x \ne 3} \right)\\
= {\left( {x - 3} \right)^2}:\left( {x - 3} \right) - {x^2} - 7x - 9\\
= x - 3 - {x^2} - 7x - 9\\
= - {x^2} - 6x - 12.\\
b)\,\,\,Khi\,\,\,x = - 1\,\,\,\,ta\,\,\,co:\,\,\,\\
A = - 2.{\left( { - 1} \right)^2} + 12.\left( { - 1} \right) = - 2 - 12 = - 14.\\
c)\,\,\,C = A + B = - 2{x^2} + 12x - {x^2} - 6x - 12\\
= - 3{x^2} + 6x - 12\\
= - 3\left( {{x^2} - 2x} \right) - 12\\
= - 3\left( {{x^2} - 2x + 1 - 1} \right) - 12\\
= - 3{\left( {x - 1} \right)^2} + 3 - 12\\
= - 3{\left( {x - 1} \right)^2} - 9\\
Vi\,\,\,{\left( {x - 1} \right)^2} \ge 0\,\,\,\forall x \ne 3\\
\Rightarrow - 3{\left( {x - 1} \right)^2} \le 0\,\,\,\forall x \ne 3\\
\Rightarrow - 3{\left( {x - 1} \right)^2} - 9 < 0\,\,\,\forall x \ne 3.\\
\Rightarrow C < 0\,\,\forall x \ne 3.
\end{array}\)
