Đáp án: $x \in {\rm{\{ }} - 1;1;2;4;5;7\} $
Giải thích các bước giải:
$\begin{array}{l}
B = \frac{{4{x^3} - 3{x^2} + 2x - 83}}{{x - 3}}\left( {dkxd:x \ne 3} \right)\\
= \frac{{4{x^3} - 12{x^2} + 9{x^2} - 27x + 29x - 87 + 4}}{{x - 3}}\\
= \frac{{4{x^2}\left( {x - 3} \right) + 9x\left( {x - 3} \right) + 29\left( {x - 3} \right) + 4}}{{x - 3}}\\
= \frac{{\left( {x - 3} \right)\left( {4{x^2} + 9x + 29} \right) + 4}}{{x - 3}}\\
= 4{x^2} + 9x + 29 + \frac{4}{{x - 3}}\\
Để:B \in Z\\
\Rightarrow \frac{4}{{x - 3}} \in Z\\
\Rightarrow \left( {x - 3} \right) \in Ư\left( 4 \right) = {\rm{\{ }} - 4; - 2; - 1;1;2;4\} \\
\Rightarrow x \in {\rm{\{ }} - 1;1;2;4;5;7\} \left( {tmdk} \right)
\end{array}$
