Đáp án:
(x;y)=(0;3); (-2;-2); (1;1); (-3;0); (2;0); (-4;1); (5;-2); (-7;3)
Giải thích các bước giải:
$x^{2}+2xy+2y+x=6$
⇔$(x^{2}+2xy)+(2y+x)=6$
⇔$x(x+2y)+(x+2y)=6$
⇔$(x+1)(x+2y)=6$
⇒x+1;x+2y∈UC(6}={-1;1;-2;2;-3;3;-6;6}
⇒(1)$\left \{ {{x+1=1} \atop {x+2y=6}} \right.$⇔ $\left \{ {{x=0} \atop {y=3}} \right.$
⇒(2)$\left \{ {{x+1=-1} \atop {x+2y=-6}} \right.$⇔ $\left \{ {{x=-2} \atop {y=-2}} \right.$
⇒(3)$\left \{ {{x+1=2} \atop {x+2y=3}} \right.$⇔ $\left \{ {{x=1} \atop {y=1}} \right.$
⇒(4)$\left \{ {{x+1=-2} \atop {x+2y=-3}} \right.$⇔ $\left \{ {{x=-3} \atop {y=0}} \right.$
⇒(5)$\left \{ {{x+1=3} \atop {x+2y=2}} \right.$⇔ $\left \{ {{x=2} \atop {y=0}} \right.$
⇒(6)$\left \{ {{x+1=-3} \atop {x+2y=-2}} \right.$⇔ $\left \{ {{x=-4} \atop {y=1}} \right.$
⇒(7)$\left \{ {{x+1=6} \atop {x+2y=1}} \right.$⇔ $\left \{ {{x=5} \atop {y=-2}} \right.$
⇒(8)$\left \{ {{x+1=-6} \atop {x+2y=-1}} \right.$⇔ $\left \{ {{x=-7} \atop {y=3}} \right.$
⇒(x;y)=(0;3); (-2;-2); (1;1); (-3;0); (2;0); (-4;1); (5;-2); (-7;3)