Đáp án:
Giải thích các bước giải:
Đặt $A=\frac{4n+8}{2n+3}(ĐKXĐ:x\neq\frac{-3}{2}$
$A=\frac{4n+6+2}{2n+3}$
$A=\frac{2(2n+3)+2}{2n+3}$
$A=2+\frac{2}{2n+3}$
$A∈Z⇔2+\frac{2}{2n+3}$
$Mà: n∈Z⇒2n+3$
$⇒2⋮2n+3$
$⇔2n+3∈Ư(2)=\{±1;±2\}$
$TH1:$
\(\left[ \begin{array}{l}2n+3=1\\2n+3=2\end{array} \right.⇔ \left[ \begin{array}{l}n=-1(TMĐK)\\n=\frac{-1}{2}(TMĐK)\end{array} \right.\)
$TH2:$
\(\left[ \begin{array}{l}2n+3=-1\\2n+3=-2\end{array} \right.⇔ \left[ \begin{array}{l}n=-2(TMĐK)\\n=\frac{-5}{2}(TMĐK)\end{array} \right.\)
Vậy $x∈\{-1;\frac{-1}{2};-2;\frac{-5}{2}\} $thì A∈Z
