Ta có: x+xy+y=9 (x,y thuộc Z)
<=>x+xy+y+1=10
=>(x+xy)+(y+1)=10
=>x.(y+1)+(y+1)=10
=>(y+1)(x+1)=10=1.10=10.1=(-1).(-10)=(-10).(-1)=2.5=5.2=(-2).(-5)=(-5).(-2)
Ta có bảng sau:
| (x+1);(y+1) | 1;10 | 10;1 | -10;-1 | -1;-10 | 2;5 | 5;2 | -2;-5 | -5;-2 |
| x;y | 0;9 | 9;0 | -11;-2 | -2;-11 | 1;4 | 4;1 | -3;-6 | -6;-3 |
Ta có :
\(x+xy+y=9\)
\(\Leftrightarrow\)\(\left(xy+x\right)+\left(y+1\right)=9+1\) ( cộng hai vế cho 1 )
\(\Leftrightarrow\)\(x\left(y+1\right)+\left(y+1\right)=10\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(y+1\right)=10\) ( đặt nhân tử chung y + 1 )
\(\Rightarrow\)\(\left(x+1\right);\left(y+1\right)\inƯ\left(10\right)\)
Ta có bảng :
| \(x+1\) | \(1\) | \(10\) | \(-1\) | \(-10\) | \(2\) | \(5\) | \(-2\) | \(-5\) |
| \(y+1\) | \(10\) | \(1\) | \(-10\) | \(-1\) | \(5\) | \(2\) | \(-5\) | \(-2\) |
| \(x\) | \(0\) | \(9\) | \(-2\) | \(-11\) | \(1\) | \(4\) | \(-3\) | \(-6\) |
| \(y\) | \(9\) | \(0\) | \(-11\) | \(-2\) | \(4\) | \(1\) | \(-6\) | \(-3\) |
Vậy \(\left(x,y\right)=\left\{\left(0;9\right),\left(9;0\right),\left(-2;-11\right),\left(-11;-2\right),\left(1;4\right),\left(4;1\right),\left(-3;-6\right),\left(-6;-3\right)\right\}\)
Chúc bạn học tốt ~