Đáp án:
\(x \in \left\{ { - 22, - 10, - 6, - 4, - 2, - 1,0,1,3,4,5,6,8,10,14,26} \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
M = \frac{{{x^2} + 4x + 12}}{{x - 2}}\\
M = \frac{{{x^2} - 2x + 6x - 12 + 24}}{{x - 2}}\\
M = x + 6 + \frac{{24}}{{x - 2}}
\end{array}\)
Vì x∈Z⇒x-2∈Z
Để M∈Z⇒
\(\begin{array}{l}
\frac{{24}}{{x - 2}} \in Z\\
\Rightarrow 24 \vdots (x - 2)\\
\Rightarrow x - 2 \in U(24) = \left\{ { - 24, - 12, - 8, - 6, - 4, - 3, - 2, - 1,1,2,3,4,6,8,12,24} \right\}\\
\Rightarrow x \in \left\{ { - 22, - 10, - 6, - 4, - 2, - 1,0,1,3,4,5,6,8,10,14,26} \right\}
\end{array}\)