Đáp án:`(x,y) in {(1,0),(3,-2),(7,-6),(5,-4)}`.
Giải thích các bước giải:
`x^2+2y^2+3xy+2x+2y-3=0`
`<=>x^2+xy+2xy+2y^2+2(x+y)-3=0`
`<=>x(x+y)+2y(x+y)+2(x+y)-3=0`
`<=>(x+y)(x+2y+2)=3`
Vì `x,y in ZZ=>x+y,x+2y+2 in ZZ`
`=>x+y,x+2y+2 in Ư(3)={+-1,+-3}`
Ta có bảng sau:
$\begin{array}{|c|c|c|}\hline x+y& 1&3&-1&-3\\\hline x+2y+2&3&1&-3&-1 \\\hline x&1 &3&7&5\\\hline y&0&-2&-6&-4 \\\hline KL&TM&TM&TM&TM\\\hline\end{array}$
Vậy `(x,y) in {(1,0),(3,-2),(7,-6),(5,-4)}`.
