Đề bài là tìm x nguyên để giá trị biểu thức nguyên

 Để `C=(x-2)/(3+2x)` nguyên

`=>x-2vdots3+2x`

`=>2(x-2)vdots3+2x`

`=>2x-4vdots3+2x`

`=>3+2x-7vdots3+2x`

Vì `3+2xvdots3+2x`

`=>7vdots3+2x`

Vì `x in ZZ`

`=>3+2x in Ư(7)={+-1;+-7}`

Ta có bảng sau:

$$\begin{array}{|c|c|c|c|c|}\hline \text{$3+2x$}&\text{1}&\text{$-1$}&\text{7}&\text{$-7$}\\\hline \text{$x$}&\text{$-1$}&\text{$-2$}&\text{2}&\text{$-5$}\\\hline \text{$C$}&\text{$-3$(t/m)}&\text{4(t/m)}&\text{0(t/m)}&\text{1(t/m)}\\\hline\end{array}$$

Vậy `x in {-1;-2;2;-5}`