Đáp án:
Giải thích các bước giải:
ta co x^2+2y^2+3xy+3x+5y=15
[x^2+3x(y+1)+9/4(y+1)^2] -9/4(y^2+2y+1)+2y^2+5y-15=0
(x+3/2y+3/2)^2-9/4y^2-9/4y-9/4+2y^2+5y-15=0
(x+3/2y+3/2)^2-1/4y^2+1/2y-69/4=0
(x+3/2y+3/2)^2-(1/4y^2-1/2y+1/4)=17
(x+3/2y+3/2)^2-(1/2y+1/2)^2=17
(x+y+1)(x+2y+2)=17
suy ra x+y+1; x+2y+2 thuoc uoc cua 17
vay (x;y)=(33;-17);(-33;15);(-15;15);(15;-17)