Đáp án+Giải thích các bước giải:
Bài `6:`
`a) xy + 2x - y =5`
`<=> x(y+2) - (y + 2) =5 -2`
`<=> (x-1)(y+2) =3`
Vì `x, y in ZZ` nên từ phương trình đã cho:
`=> (x-1) in Ư(3)`
`=> (x-1) in {±1; ±3}`
Ta có bảng sau:
\begin{array}{|c|c|c|}\hline \text{x-1}&\text{-1}&\text{1}&\text{-3}&\text{3}\\\hline \text{x}&\text{0}&\text{2}&\text{-2}&\text{4}\\\hline \text{y+2}&\text{-3}&\text{3}&\text{-1}&\text{1}\\\hline \text{y}&\text{-5}&\text{1}&\text{-3}&\text{-1}\\\hline \text{Đối chiếu}&\text{tm}&\text{tm}&\text{tm}&\text{tm}\\\hline\end{array}
Vậy `(x; y) in {(0; -5), (2; 1), (-2; -3), (4;-1)}`
`b) 2xy - 6x + y =7`
`<=> 2x(y-3)+(y-3) = 7-3`
`<=> (2x+1)(y-3) = 4`
Vì `x, y in ZZ` nên từ phương trình đã cho:
`=> (2x+1) in Ư(4)`
`=> (2x+1) in {±1; ±2; ±4}`
Ta có bảng sau:
\begin{array}{|c|c|c|}\hline \text{2x+1}&\text{-1}&\text{1}&\text{-2}&\text{2}&\text{-4}&\text{4}\\\hline \text{x}&\text{-1}&\text{0}&\text{-1,5}&\text{0,5}&\text{-2,5}&\text{1,5}\\\hline \text{y-3}&\text{-4}&\text{4}&\text{-2}&\text{2}&\text{-1}&\text{1}\\\hline \text{y}&\text{-1}&\text{7}&\text{1}&\text{5}&\text{2}&\text{4}\\\hline \text{Đối chiếu}&\text{tm}&\text{tm}&\text{ktm}&\text{ktm}&\text{ktm}&\text{ktm}\\\hline\end{array}
Vậy `(x; y) in {(-1; -1); (0; 7)}`
