Lời giải:
Từ \(x+y+xy=0\Rightarrow x(y+1)+(y+1)=1\)
\(\Leftrightarrow (x+1)(y+1)=1\)
Do \(x,y\in\mathbb{Z}\) nên chỉ có thể xảy ra 2TH:
TH1: \(\left\{\begin{matrix} x+1=1\\ y+1=1\end{matrix}\right.\Rightarrow x=y=0\)
TH2: \(\left\{\begin{matrix} x+1=-1\\ y+1=-1\end{matrix}\right.\Rightarrow x=y=-2\)
Vậy \((x,y)=(0,0);(-2,-2)\)
\(x+y+xy=0\)
\(x+y\left(1+x\right)=0\)
\(1+x+y\left(1+x\right)=0+1\)
\(\left(1+x\right)+y\left(1+x\right)=1\)
\(\left(1+x\right)\left(1+y\right)=1\)
Vì \(x;y\in Z\Leftrightarrow1+x;1+y\in Z;1+x;1+y\inƯ\left(1\right)\)
Sau đó lập bảng là ok thoy!
\(x+y+xy=0\)
\(\Rightarrow x.\left(y+1\right)+y+1=1\)
\(\Rightarrow\left(y+1\right).\left(x+1\right)=1\)
\(\Rightarrow x+1;y+1\inƯ\left(1\right)\)
\(\Rightarrow x+1;y+1\in\left\{-1;1\right\}\)
Ta có bảng sau:
| \(x+1\) | -1 | 1 |
| \(y+1\) | -1 | 1 |
| x | -2 | 0 |
| y | -2 | 0 |
| Chọn or loại | Chọn | Chọn |
Vậy........
Chúc bạn học tốt!!!
\(x+y+xy=0\)
\(\Rightarrow x+y+xy+1=1\)
\(\Rightarrow x\left(y+1\right)+1\left(y+1\right)=1\)
\(\left(x+1\right)\left(y+1\right)=1\)
\(\Rightarrow x+1;y+1\inƯ\left(1\right)\)
\(Ư\left(1\right)=\left\{1;-1\right\}\)
\(\Rightarrow x+1=1\Rightarrow x=0\)
\(y+1=-1\Rightarrow y=-2\)
\(\Rightarrow x+1=-1\Rightarrow x=-2\)
\(y+1=1\Rightarrow y=0\)
