Question

Tìm các số nguyên x,y thỏa mãn pt sau:

\(2x^2+y^2+3xy+3x+2y+2=0\)

Answer 1

2x² + y² + 3xy + 3x + 2y + 2 = 0

<=> 16x² + 8y² + 24xy + 24x + 16y + 16 = 0

<=> (4x)² + 24x(y+1) + 8y² + 16y + 16 = 0

<=> (4x)² + 24x(y+1) + [3(y + 1)]² - [3(y + 1)]² + 8y² + 16y + 16 = 0

<=> (4x + 3y + 3)² - 9y² - 18y - 9 + 8y² + 16y + 16 = 0

<=> (4x + 3y + 3)² - y² - 2y - 1 + 8 = 0

<=> (4x + 3y + 3)² - (y + 1)² = - 8

<=> (y + 1)² - (4x + 3y + 3)² = 8

<=> (y + 1 +4x + 3y + 3)(y + 1 - 4x - 3y - 3) = 8

<=> 4(x + y + 4)( - 4x - 2y - 2) = 8

<=> (x + y + 4)( 2x + y + 1) = -1

<=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y+4=-1\\2x+y+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}x+y+4=1\\2x+y+1=-1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=-4\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\end{matrix}\right.\)

vậy nghiệm (x;y) = (2 ; - 4) (-2; 2)