Đáp án:
`↓↓`
Giải thích các bước giải:
`x^2+2y^2+2xy+2x+6y+1=0`
`=> (x^2+y^2+1+2xy+2y+2x)+(y^2+4y+4)-4==0`
`=> (x+y+1)^2+(y+2)^2=4`
Mà `(x+y+1)^2; (y+2)^2` là số chính phương
`=> ` $\left\{\begin{matrix}\left[ \begin{array}{l}(x+y+1)^2=4\\(y+2)^2=0\end{array} \right. & \\\left[ \begin{array}{l}(x+y+1)^2=0\\(y+2)^2=4\end{array} \right.& \end{matrix}\right.$
`TH_1:`
$\left\{\begin{matrix}(x+y+1)^2=4& \\(y+2)^2=0& \end{matrix}\right.$
`=> `$\left\{\begin{matrix}\left[ \begin{array}{l}x+y+1=2\\x+y+1=-2\end{array} \right.& \\y+2=0& \end{matrix}\right.$ `=>` $\left\{\begin{matrix}\left[ \begin{array}{l}x+y=1\\x+y=-3\end{array} \right.& \\y=-2& \end{matrix}\right.$ `=>` $\left\{\begin{matrix}\left[ \begin{array}{l}x+y=1\\x+y=-3\end{array} \right.& \\y=-2& \end{matrix}\right.$
`=>`
$\left\{\begin{matrix}x=3& \\y=-2& \end{matrix}\right.$
$\left\{\begin{matrix}x=-1& \\y=-2& \end{matrix}\right.$
`TH_2:`
$\left\{\begin{matrix}(x+y+1)^2=0& \\(y+2)^2=4& \end{matrix}\right.$
`=> `$\left\{\begin{matrix}x+y+1=0& \\\left[ \begin{array}{l}y+2=2\\y+2=-2\end{array} \right.& \end{matrix}\right.$ `=> `$\left\{\begin{matrix}x+y=-1& \\\left[ \begin{array}{l}y=0\\y=-4\end{array} \right.& \end{matrix}\right.$
`=>`
$\left\{\begin{matrix}x=-1& \\y=0& \end{matrix}\right.$
$\left\{\begin{matrix}x=3& \\y=-4& \end{matrix}\right.$
Vậy `(x;y) in { (3; -2); (-1; -2); (-1; 0); (3; -4) }`
