Đáp án:
$\left( {a;b} \right) = \left( {0; - 1} \right)$
Giải thích các bước giải:
Ta có:
Đặt $P\left( x \right) = {x^4} + ax + b$
Để $P\left( x \right) \vdots \left( {{x^2} - 1} \right)$
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
P\left( 1 \right) = 0\\
P\left( { - 1} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{1^4} + a + b = 0\\
{\left( { - 1} \right)^4} - a + b = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a + b = - 1\\
- a + b = - 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {a + b} \right) + \left( { - a + b} \right) = - 2\\
a + b = - 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2b = - 2\\
a + b = - 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = - 1\\
a = 0
\end{array} \right.
\end{array}$
Vậy $\left( {a;b} \right) = \left( {0; - 1} \right)$
