Giải thích các bước giải:
\(\begin{array}{l}
a,\\
P = \frac{{{x^4} - 3{x^2} + 5}}{{x - 3}}\,\,\,\,\,\,\left( {x \ne 3} \right)\\
= \frac{{\left( {{x^4} - 3{x^3}} \right) + \left( {3{x^3} - 9{x^2}} \right) + \left( {6{x^2} - 18x} \right) + \left( {18x - 54} \right) + 59}}{{x - 3}}\\
= \frac{{\left( {x - 3} \right)\left( {{x^3} + 3{x^2} + 6x + 18} \right) + 59}}{{x - 3}}\\
= \left( {{x^3} + 3{x^2} + 6x + 18} \right) + \frac{{59}}{{x - 3}}\\
x \in Z \Rightarrow \left( {{x^3} + 3{x^2} + 6x + 18} \right) \in Z\\
P \in Z \Leftrightarrow \frac{{59}}{{x - 3}} \in Z \Rightarrow \left( {x - 3} \right) \in \left\{ { \pm 1;\,\,\, \pm 59} \right\}\\
\Rightarrow x \in \left\{ { - 56;\,\,2;\,\,4;\,\,62} \right\}\\
b,\\
Q = \frac{{2{x^3} + {x^2} + 2x + 13}}{{2x - 1}} = \frac{{\left( {2{x^3} - {x^2}} \right) + \left( {2{x^2} - x} \right) + \left( {2x - 1} \right) + x + 14}}{{2x - 1}}\\
= \frac{{\left( {2x - 1} \right)\left( {{x^2} + x + 1} \right) + x + 14}}{{2x - 1}}\\
= \left( {{x^2} + x + 1} \right) + \frac{{x + 14}}{{2x - 1}}\\
x \in Z \Rightarrow \left( {{x^2} + x + 1} \right) \in Z\\
P \in Z \Leftrightarrow \frac{{x + 14}}{{2x - 1}} \in Z\\
\Rightarrow 2.\frac{{x + 14}}{{2x - 1}} \in Z\\
\Leftrightarrow \frac{{2x + 28}}{{2x - 1}} \in Z\\
\Leftrightarrow \frac{{\left( {2x - 1} \right) + 29}}{{2x - 1}} \in Z\\
\Leftrightarrow 1 + \frac{{29}}{{2x - 1}} \in Z\\
\Rightarrow \frac{{29}}{{2x - 1}} \in Z \Rightarrow \left( {2x - 1} \right) \in \left\{ { \pm 1;\,\, \pm 29} \right\}\\
\Rightarrow x \in \left\{ { - 14;\,\,0;\,\,1;\,\,15} \right\}\,\,\,\,\,\,\left( {t/m} \right)
\end{array}\)
