`xy+2x+2y=-16`
`<=>x(y+2)+2y+4=-12`
`<=>x(y+2)+2(y+2)=-12`
`<=>(x+2)(y+2)=-12`
Vì `x,y in ZZ=>x+2,y+2 in ZZ`
`=>x+2,y+2 in Ư(-12)={+-1,+-2,+-3,+-4,+-6,+-12}`
Đến đấy chia th rồi giải thui :v
xy+2x+2y=−16xy+2x+2y=-16
⇔x(y+2)+2y+4=−12⇔x(y+2)+2y+4=-12
⇔x(y+2)+2(y+2)=−12⇔x(y+2)+2(y+2)=-12
⇔(x+2)(y+2)=−12⇔(x+2)(y+2)=-12
Vì x,y∈Z⇒x+2,y+2∈Zx,y∈ℤ⇒x+2,y+2∈ℤ
⇒x+2,y+2∈Ư(−12)={±1,±2,±3,±4,±6,±12}
xy+2x+2y+4=-16+4
x.(y+2)+2.(y+2)=-12
(x+2).(y+2)=-12
Ta có: -12=-1.12=12.-1=1.-12=-12.1=-6.2=2.-6=6.-2=-2.6=-3.4=3.-4=4.-3=-4.3
⇒(x,y)=(-3,10);(10,-3);(-1,-14);(-14,1);(1,-14);(-14,1);(4,-4),(-4,4);(-5,2);(1,-6);(2,-5);(-6,1)
xy+2x+2y=−16xy+2x+2y=-16
⇔x(y+2)+2y+4=−12⇔x(y+2)+2y+4=-12
⇔x(y+2)+2(y+2)=−12⇔x(y+2)+2(y+2)=-12
⇔(x+2)(y+2)=−12⇔(x+2)(y+2)=-12
Vì x,y∈Z⇒x+2,y+2∈Zx,y∈ℤ⇒x+2,y+2∈ℤ
⇒x+2,y+2∈Ư(−12)={±1,±2,±3,±4,±6,±12}
\(xy+2x+2y=-16\)
\(\Rightarrow x.\left(y+2\right)+2y+4=-12\)
\(\Rightarrow x.\left(y+2\right)+2.\left(y+2\right)=-12\)
\(\Rightarrow\left(x+2\right).\left(y+2\right)=-12\)
\(\Rightarrow\left(x+2\right)\) và \(\left(y+2\right)\inƯ\left(-12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
x+2 | -12 | -6 | -4 | -3 | -2 | -1 | 1 | 2 | 3 | 4 | 6 | 12 |
| y+2 | 1 | 2 | 3 | 4 | 6 | 12 | -12 | -6 | -4 | -3 | -2 | -1 |
| x | -14 | -8 | -6 | -5 | -4 | -3 | -1 | 0 | 1 | 2 | 4 | 10 |
| y | -1 | 0 | 1 | 2 | 4 | 10 | -14 | -8 | -6 | -5 | -4 | -3 |
Vậy \(\left(x;y\right)=\left\{\left(-14;-1\right);\left(-8;0\right);\left(-6;1\right);\left(-5;2\right);\left(-4;4\right);\left(-3;10\right);\left(-1;-14\right);\left(0;-8\right);\left(1;-6\right);\left(2;-5\right);\left(4;-4\right);\left(10;3\right)\right\}\)
