Đáp án:
$\left( {a;b} \right) = \left( {5;6} \right)$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
{x^4} - {x^3} + b{x^2} - x + a\\
= {x^2}\left( {{x^2} - x + 5} \right) + \left( {b - 5} \right)\left( {{x^2} - x + 5} \right) + \left( {b - 6} \right)x + a - 5\left( {b - 5} \right)\\
= \left( {{x^2} - x + 5} \right)\left( {{x^2} + b - 5} \right) + \left( {b - 6} \right)x + a - 5\left( {b - 5} \right)
\end{array}$
Để $\left( {{x^4} - {x^3} + b{x^2} - x + a} \right) \vdots \left( {{x^2} - x + 5} \right)$
$\begin{array}{l}
\Leftrightarrow \left( {\left( {{x^2} - x + 5} \right)\left( {{x^2} + b - 5} \right) + \left( {b - 6} \right)x + a - 5\left( {b - 5} \right)} \right) \vdots \left( {{x^2} - x + 5} \right)\\
\Leftrightarrow \left( {\left( {b - 6} \right)x + a - 5\left( {b - 5} \right)} \right) \vdots \left( {{x^2} - x + 5} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
b - 6 = 0\\
a - 5\left( {b - 5} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = 6\\
a = 5\left( {b - 5} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = 6\\
a = 5
\end{array} \right.\\
\Rightarrow \left( {a;b} \right) = \left( {5;6} \right)
\end{array}$
Vậy $\left( {a;b} \right) = \left( {5;6} \right)$
