Đáp án:
$\begin{array}{l}1)\\ - 3{x^3}\left( {2a.{x^2} - bx + c} \right)\\ = - 6a.{x^5} + 3b.{x^4} - 3c.{x^3}\\ = - 6{x^5} + 9{x^4} - 3{x^3}\\ \Leftrightarrow \left\{ \begin{array}{l} - 6a = - 6\\3b = 9\\ - 3c = - 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a = 1\\b = 3\\c = 1\end{array} \right.\\Vậy\,a = 1;b = 3;c = 1\\2)\\\left( {a.x + b} \right)\left( {{x^2} - c.x + 2} \right)\\ = a.{x^3} - a.c.{x^2} + 2a.x + b.{x^2} - bc.x + 2b\\ = a.{x^3} + \left( { - ac + b} \right).{x^2} + \left( {2a - bc} \right).x + 2b\\ = {x^3} + {x^2} - 2\\ \Leftrightarrow \left\{ \begin{array}{l}a = 1\\ - a.c + b = 1\\2a - bc = 0\\2b = - 2\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a = 1\\ - c + b = 1\\2 - bc = 0\\b = - 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a = 1;b = - 1\\c = b - 1 = - 2\\bc = - 2\end{array} \right.\\Vậy\,a = 1;b = - 1;c = - 2\end{array}$
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