\(xy+y+x+1=5\)
\(\Leftrightarrow y\left(x+1\right)+\left(x+1\right)=5\)
\(\Leftrightarrow\left(y+1\right)\left(x+1\right)=5\)
=> y + 1 và x + 1 \(\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng:
| y+1 | -5 | 5 | -1 | 1 |
| x+1 | -1 | 1 | -5 | 5 |
| y | -6 | 4 | -2 | 0 |
| x | -2 | 0 | -6 | 4 |
Vậy các cặp (x;y) là (-2;-6) ; (0;4) ; (-6;-2) ; (4;0)
\(xy-y+x-1=7\)
\(\Leftrightarrow y\left(x-1\right)+\left(x-1\right)=7\)
\(\Leftrightarrow\left(y+1\right)\left(x-1\right)=7\)
=> y + 1 và x - 1 \(\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
Ta có bảng:
| y+1 | -7 | 7 | -1 | 1 |
| x-1 | -1 | 1 | -7 | 7 |
| y | -8 | 6 | -2 | 0 |
| x | 0 | 2 | -6 | 8 |
Vậy các cặp (x;y) là (0;-8) ; (2;6) ; (-6;-2) ; (8;0)
a, \(x.y+y+x+1=5\Leftrightarrow x\left(y+1\right)+x+1=5\)
\(\Leftrightarrow x\left(y+1\right)+x=4\Leftrightarrow x\left(y+2\right)=4\)
\(\Rightarrow x;y+2\inƯ\left(4\right)\Rightarrow x;y+2\in\left\{\pm1;\pm2;\pm4\right\}\)
| x | 1 | 2 | 4 | -1 | -2 | -4 |
| y + 2 | 4 | 2 | 1 | -4 | -2 | -1 |
| y | 2 | 0 | -1 | -6 | -4 | -3 |
Vậy các cặp số nguyên (x;y) thỏa mãn là (1;2);(2;0);(4;-1);(-1;-6);(-2;-4);(-4;-3)
b, \(x.y-y+x-1=7\Leftrightarrow x\left(y-1\right)+x-1=7\)
\(\Leftrightarrow x\left(y-1\right)+x=8\Leftrightarrow x.y=8\)
\(\Rightarrow x;y\inƯ\left(8\right)\Rightarrow x;y\in\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
Vậy ...
