$\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l} c – 3 = 0\\ a = d – 3c + 2\\ 2c – 3d = 0\\ b = 2d \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} c = 3\\ d = 2\\ a = d – 3c + 2\\ b = 2d \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} c = 3\\ d = 2\\ a = – 5\\ b = 4 \end{array} \right. \end{array}$
Đáp án:
$\left( {a;b;c;d} \right) = \left( { – 5;4;3;2} \right)$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
{x^4} + a{x^2} + b = \left( {{x^2} – 3x + 2} \right)\left( {{x^2} + cx + d} \right)\\
\Leftrightarrow {x^4} + a{x^2} + b = {x^4} + \left( {c – 3} \right){x^3} + \left( {d – 3c + 2} \right){x^2} + \left( {2c – 3d} \right)x + 2d
\end{array}$
Đồng nhất hệ số hai vế ta có:
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
c – 3 = 0\\
a = d – 3c + 2\\
2c – 3d = 0\\
b = 2d
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
c = 3\\
d = 2\\
a = d – 3c + 2\\
b = 2d
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
c = 3\\
d = 2\\
a = – 5\\
b = 4
\end{array} \right.
\end{array}$
Vậy $\left( {a;b;c;d} \right) = \left( { – 5;4;3;2} \right)$
.