a/ ĐK: $x\ne 2$
$A∈\Bbb Z\\→\dfrac{3}{x-2}∈\Bbb Z\\→3\vdots x-2\\→x-2∈Ư(3)=\{±1;±3\}$
Ta có bảng:
$\begin{array}{|c|c|c|}\hline x-2&1&-1&3&-3\\\hline x&3&1&5&-1\\\hline\end{array}$
mà $x\ne 2$
$→x\in\{3;1;5;-1\}$
Vậy $x\in\{3;1;5;-1\}$ thì $A\in\Bbb Z$
b/ ĐK: $x\ne -\dfrac{1}{3}$
$\dfrac{3x^2-2x+1}{3x+1}\\=\dfrac{3x^2+x-3x+1}{3x+1}\\=\dfrac{x(3x+1)-3x+1}{3x+1}\\=x-\dfrac{3x-1}{3x+1}\\=x-\dfrac{3x+1-2}{3x+1}\\=x-1-\dfrac{2}{3x+1}$
$\dfrac{3x^2-2x+1}{3x+1}\in\Bbb Z\\→x-1-\dfrac{2}{3x+1}\in\Bbb Z\\→\begin{cases}x\in\Bbb Z\\2\vdots 3x+1\end{cases}$
$2\vdots 3x+1\\→3x+1∈Ư(2)=\{±1;±2\}$
Ta có bảng:
$\begin{array}{|c|c|c|}\hline 3x+1&1&-1&2&-2\\\hline 3x&0&-2&1&-3\\\hline x&0&-\dfrac{2}{3}&\dfrac{1}{3}&-1\\\hline\end{array}$
mà $x\ne -\dfrac{1}{3},x\in\Bbb Z$
$→x∈\{0;-1\}$
Vậy $x∈\{0;-1\}$ thì $\dfrac{3x^2-2x+1}{3x+1}\in\Bbb Z$
