Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
xy - {x^2} + x - y\\
= \left( {xy - {x^2}} \right) + \left( {x - y} \right)\\
= x.\left( {y - x} \right) + \left( {x - y} \right)\\
= - x.\left( {x - y} \right) + \left( {x - y} \right)\\
= \left( {x - y} \right).\left( { - x + 1} \right)\\
b,\\
{x^4} - 3{x^3} - 6x + a\\
= \left( {{x^4} - 3{x^3} - 2{x^2}} \right) + \left( {2{x^2} - 6x - 4} \right) + \left( {a + 4} \right)\\
= {x^2}.\left( {{x^2} - 3x - 2} \right) + 2.\left( {{x^2} - 3x - 2} \right) + \left( {a + 4} \right)\\
= \left( {{x^2} - 3x - 2} \right)\left( {{x^2} + 2} \right) + \left( {a + 4} \right)\\
\left( {{x^4} - 3{x^3} - 6x + a} \right)\,\, \vdots \,\,\left( {{x^2} - 3x - 2} \right)\\
\Leftrightarrow \left[ {\left( {{x^2} - 3x - 2} \right)\left( {{x^2} + 2} \right) + \left( {a + 4} \right)} \right]\,\, \vdots \,\,\left( {{x^2} - 3x - 2} \right)\\
\Rightarrow \left( {a + 4} \right)\,\, \vdots \,\,\left( {{x^2} - 3x - 2} \right)\\
\Rightarrow a + 4 = 0\\
\Leftrightarrow a = - 4
\end{array}\)
