Đáp án:
b. \(\left\{ \begin{array}{l}
a = 1\\
c = 0\\
b = - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.({x^2} + cx + 2)(ax + b)\\
= a{x^3} + b{x^2} + ac{x^2} + bcx + 2ax + 2b\\
= a{x^3} + \left( {b + ac} \right){x^2} + \left( {bc + 2a} \right)x + 2b\\
Do:({x^2} + cx + 2)(ax + b) = {x^3} - {x^2} + 2\\
\to a{x^3} + \left( {b + ac} \right){x^2} + \left( {bc + 2a} \right)x + 2b = {x^3} - {x^2} + 2\\
\to \left\{ \begin{array}{l}
a = 1\\
b + ac = - 1\\
bc + 2a = 0\\
2b = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
a = 1\\
b = 1\\
c = - 2
\end{array} \right.\\
b.(a.{y^2} + by + c)(y + 3)\\
= a{y^3} + 3a{y^2} + b{y^2} + 3by + cy + 3c\\
= a{y^3} + \left( {3a + b} \right){y^2} + \left( {3b + c} \right)y + 3c\\
Do:(a.{y^2} + by + c)(y + 3) = {y^3} + 2.{y^2} - 3y\\
\to a{y^3} + \left( {3a + b} \right){y^2} + \left( {3b + c} \right)y + 3c = {y^3} + 2.{y^2} - 3y\\
\to \left\{ \begin{array}{l}
a = 1\\
3a + b = 2\\
3b + c = - 3\\
3c = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
a = 1\\
c = 0\\
b = - 1
\end{array} \right.
\end{array}\)
