Đáp án: $x \in \left\{ { - 10; - 4; - 2;0;2;8} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
B = \dfrac{{{x^2} - 2x + 6}}{{x + 1}}\\
= \dfrac{{{x^2} + x - 3x - 3 + 9}}{{x + 1}}\\
= \dfrac{{x\left( {x + 1} \right) - 3\left( {x + 1} \right) + 9}}{{x + 1}}\\
= \dfrac{{\left( {x + 1} \right)\left( {x - 3} \right) + 9}}{{x + 1}}\\
= x - 3 + \dfrac{9}{{x + 1}}\\
B \in Z\\
\Leftrightarrow \dfrac{9}{{x + 1}} \in Z\\
\Leftrightarrow \left( {x + 1} \right) \in \left\{ { - 9; - 3; - 1;1;3;9} \right\}\\
\Leftrightarrow x \in \left\{ { - 10; - 4; - 2;0;2;8} \right\}\\
Vậy\,x \in \left\{ { - 10; - 4; - 2;0;2;8} \right\}
\end{array}$
