Đáp án:
$ (x,y)\in\{(-1,0),(3,-4),(3,-2),(1,-2)\}$
Giải thích các bước giải:
Ta có :
$x^2+2y^2+2xy+2x+6y+1=0$
$\rightarrow x^2+2x(y+1)+(y+1)^2+y^2+4y=0$
$\rightarrow (x+y+1)^2+(y^2+4y+4)=4$
$\rightarrow (x+y+1)^2+(y+2)^2=4$
$\rightarrow ((x+y+1)^2,(y+2)^2)\in\{(0,4),(4,0)\}$
$\rightarrow (x+y+1,y+2)\in\{(0,2),(0,-2),(2,0),(-2,0)\}$
$\rightarrow (x+y,y)\in\{(-1,0),(-1,-4),(1,-2),(-3,-2)\}$
$\rightarrow (x,y)\in\{(-1,0),(3,-4),(3,-2),(1,-2)\}$
