b) \(xy+x+y=0\)
\(\Leftrightarrow x\left(y+1\right)+y=0\)
\(\Leftrightarrow x\left(y+1\right)+y+1=1\)
\(\Leftrightarrow\left(y+1\right)\left(x+1\right)=1\)
làm nốt
\(a,\text{ }\left(x+4\right)\left(y+3\right)=3\)
\(\Rightarrow\text{ }\left(x+4\right),\text{ }\left(y+3\right)\inƯ\left(3\right)\)
Ta có bảng :
| x + 4 | - 1 | 1 | - 3 | 3 |
| y + 3 | - 3 | 3 | - 1 | 1 |
| x | - 5 | - 3 | - 7 | - 1 |
| y | - 6 | 0 | - 4 | - 2 |
\(\Rightarrow\text{ }\left(x,y\right)=\left(-5\text{ ; }-6\right),\left(-3\text{ ; }0\right)\text{ , }\left(-7\text{ ; }-4\right),\left(-1\text{ ; }-2\right)\)
Câu b giải tiếp bài anh Lê Tài Bảo Châu
\(\Rightarrow\text{ }\left(y+1\right),\left(x+1\right)\inƯ\left(1\right)\)
Ta có bảng :
| x + 1 | - 1 | 1 |
| y + 1 | - 1 | 1 |
| x | - 2 | 0 |
| y | - 2 | 0 |
\(\Rightarrow\text{ }\left(x,y\right)=\left(-2\text{ , }0\right),\text{ }\left(0\text{ , }0\right)\)
