cho x,y là các số thực thỏa mãn x^2 ± 2y^2 + 2xy + 2x + 6y +1 = 0

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) }`