\(xy+2x+y+11=0\)
\(\Leftrightarrow x\left(y+2\right)+\left(y+2\right)=11+2\)
\(\Leftrightarrow\left(x+1\right)\left(y+2\right)=13\)
\(\Rightarrow\left(x+1\right)\) và \(\left(y+2\right)\) lần lược \(\inƯ\left(13\right)=\left\{1;-1;13;-13\right\}\)
TH 1 : \(x+1\inƯ\left(13\right)\)
+ ) x + 1 = 1
=> x = 0 ( nhận )
+ ) x + 1 = -1
=> x = -1 - 1
=> x = -2 ( nhận )
+ ) x + 1 = 13
=> x = 12 ( nhận )
+ ) x + 1 = -13
=> x = -14 ( nhận )
TH2 : \(y+2\inƯ\left(13\right)\)
+ ) y + 2 = 1
=> y = -1 ( nhận )
+ ) y + 2 = -1
=> y = -3 ( nhận )
+ ) y + 2 = 13
=> y = 11 ( nhận )
+) y + 2 = -13
=> y = -15 ( nhận )
Vậy .......
Ta có:
xy+2x+y+11=0
\(\Rightarrow\)x(y+2)+y+11=0
\(\Rightarrow\)x(y+2)+(y+2)+9=0
\(\Rightarrow\)(y+2).(x+1)+9=0
(y+2).(x+1)=0-9
(y+2).(x+1)=-9
\(\Rightarrow\)(y+2)\(\in\)Ư(-9)
Ư(-9)={1;9;-1;-9}
\(\Rightarrow\)y+2={1;9;-1;-9}
\(\Rightarrow\)y={-1;7;-3;-11}
*Khi y=-1 thì x=-10
*Khi y=7 thì x=-2
*khi y=-3 thì x=8
*khi y=-11 thì x=0
Vậy y=-1;x=-10
y=7;x=-2
y=-3;x=8
y=-11;x=0
xy+2x+y+11=0
xy+2x+y=0-11
xy+2x+y=-11
x.(y+2)+y=-11
x.(y+2)+(y+2)=(-11)+2
(x+1).(y+2)=-9
Ta có bảng sau:
| x+1 | 1 | 3 | 9 | -1 | -3 | -9 |
| y+2 | 9 | 3 | 1 | -9 | -3 | -1 |
| x | 0 | 2 | 8 | -2 | -4 | -10 |
| y | 7 | 1 | -1 | -11 | -5 | -3 |
Vậy \(\left\{{}\begin{matrix}x=0\\y=7\end{matrix}\right.\),\(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\),\(\left\{{}\begin{matrix}x=8\\y=-1\end{matrix}\right.\),\(\left\{{}\begin{matrix}x=-2\\y=-11\end{matrix}\right.\),\(\left\{{}\begin{matrix}x=-4\\y=-5\end{matrix}\right.\),\(\left\{{}\begin{matrix}x=-10\\y=3\end{matrix}\right.\)
