Câu 1:
xy+2x+2y=9
⇔xy+2x+2y+4=9+4
⇔(xy+2x)+(2y+4)=13
⇔x(y+2)+2(y+2)=13
⇔(x+2)(y+2)=13
⇒(x+2); (y+2) ∈Ư(13)={±1;±13}
Ta có bảng sau:
x -3 -1 -15 11
x+2 -1 1 -13 13
y+2 -13 13 -1 1
y -15 11 -3 -1
Vậy (x;y)=(-3;-15),(-1;11),(-15;-3),(11;-1)
xy+3x-7y=21
⇔xy+3x-7y-21=0
⇔(xy+3x)-(7y+21)=0
⇔x(y+3)-7(y+3)=0
⇔(x-7)(y+3)=0
⇔$\left \{ {{x-7=0} \atop {y+3=0}} \right.$⇔$\left \{ {{x=7} \atop {y=-3}} \right.$
Vậy (x;y)=(7;-3)
xy+3x-y=14
⇔xy+3x-y-3=14-3
⇔(xy+3x)-(y+3)=11
⇔x(y+3)-(y+3)=11
⇔(x-1)(y+3)=11
⇒(x-1); (y+3) ∈Ư(11)={±1;±11}
Ta có bảng sau:
x 0 2 -10 12
x-1 -1 1 -11 11
y+3 11 -11 1 -1
y 8 -14 -2 -4
Vậy (x;y)=(0;8),(2;-14),(-10;-2),(12;-4).
