Đáp án:
Giải thích các bước giải:
$Q=\frac{27-2x}{12-x}(ĐKXĐ:x\neq12)$
$Q∈Z⇒\frac{27-2x}{12-x}∈Z$
$Mà:x∈Z ⇒ (27-2x) ∈Z$
$⇒(27-2x) ⋮ (12-x)$
$⇔(3+24-x) ⋮ (12-x)$
$⇒[3+2(12-x)] ⋮ (12-x)$
$⇒3 ⋮ (12-x)$
$⇒ 12-x ∈ Ư(3)=\{±1;±3\}
$TH1:$
\(\left[ \begin{array}{l}12-x=1\\12-x=3\end{array} \right.⇔\left[ \begin{array}{l}x=11(TMĐK)\\x=9(TMĐK)\end{array} \right.\)
$TH2:$
\(\left[ \begin{array}{l}12-x=-1\\12-x=-3\end{array} \right.⇔\left[ \begin{array}{l}x=13(TMĐK)\\x=15(TMĐK)\end{array} \right.\)
Vậy $x∈\{11;9;13;15\} thì Q∈Z$
