1) \(xy-2x-y=-6\Rightarrow x\left(y-2\right)-y=-6\Rightarrow x\left(y-2\right)-y+2=-6+2\)
\(\Rightarrow x\left(y-2\right)-\left(y-2\right)=-4\Rightarrow\left(y-2\right)\left(x-1\right)=-4\)
\(\Rightarrow x-1\inƯ\left(-4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Ta có bảng sau:
| x - 1 | 1 | -1 | 2 | -2 | 4 | -4 |
| y - 2 | -4 | 4 | -2 | 2 | -1 | 1 |
Suy ra ta có các cặp (x,y) sau:
| x | 2 | 0 | 3 | -1 | 5 | -3 |
| y | -2 | 6 | 0 | 4 | 1 | 3 |
2) \(|x+1|+|x-2|+|x+7|=5x-1\)
Ta thấy: \(|x+1|\ge0,|x-2|\ge0,|x+7|\ge0\) với \(\forall x\inℤ\)
Mà \(|x+1|+|x-2|+|x+7|=5x-10\Rightarrow5x-10\ge0\Rightarrow5x\ge10\Rightarrow x\ge2>0\)
\(\Rightarrow|x+1|=x+1,|x-2|=x-2,|x+7|=x+7\)
\(\Rightarrow|x+1|+|x-2|+|x+7|=x+1+x-2+x+7=5x-10\)
\(\Rightarrow\left(x+x+x\right)+\left(1-2+7\right)=5x-10\Rightarrow3x+6=5x-10\)
\(\Rightarrow3x-5x=-10-6\Rightarrow-2x=-16\Rightarrow x=\frac{-16}{-2}=8\)
