Câu 5 (0,5 điểm): Tìm các số nguyên x, y biết: xy + 3x + 3y=−16

Câu `5:`

`xy+3x+3y=-16`

`=>x(y+3)+3y+9=-16+9`

`=>x(y+3)+3(y+3)=-7`

`=>(y+3)(x+3)=-7`

Ta có bảng sau:

\begin{array}{|c|c|c|}\hline \text{x+3}&\text{-7}&\text{-1}&\text{1}&\text{7}\\\hline \text{y+3}&\text{1}&\text{7}&\text{-7}&\text{-1}\\\hline \text{x}&\text{-10}&\text{-4}&\text{-2}&\text{4}\\\hline \text{y}&\text{-2}&\text{4}&\text{-10}&\text{-4}\\\hline \text{}&\text{tm}&\text{tm}&\text{tm}&\text{tm}\\\hline\end{array}

Vậy `(x;y) in {(-4;4),(4;-4),(-10;-2),(-2;-10)}`