Question

tìm x,y nguyên thỏa mãn: x² + 2y² + 3xy - x - y + 3 = 0

Answer 1

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

\(\Leftrightarrow\left(x+y\right)\left(x+2y\right)-\left(x+y\right)=-3\)

\(\Leftrightarrow\left(x+y\right)\left(x+2y-1\right)=-3\)

Ta có bảng:

x+y	-3	-1	1	3
x+2y-1	1	3	-3	-1
x	-4	-6	4	6
y	3	5	-3	-3

Vậy \(\left(x;y\right)=\left(-4;3\right);\left(-6;5\right);\left(4;-3\right);\left(6;-3\right)\)