Ta có : \(2xy+y=10x+17\)
\(\Leftrightarrow2xy+y-10x-17\Leftrightarrow y\left(2x+1\right)-10x+5=12\)
\(\Leftrightarrow y\left(2x+1\right)-5\left(2x+1\right)=12\Leftrightarrow\left(y-5\right)\left(2x+1\right)=12\)
Vì : \(y\in Z\Rightarrow y-5\in Z\)
\(x\in Z\Rightarrow2x+1\in Z\)
\(\Rightarrow y-5;2x+1\inƯ\left(12\right)\)
Mà : \(x\in Z\Rightarrow2x+1\) là số lẻ
Ta có bảng sau :
| 2x + 1 | 1 | 3 | -1 | -3 |
| y - 5 | 12 | 4 | -12 | -4 |
| x | 0 | 2 | -1 | -2 |
| y | 17 | 9 | -7 | 1 |
Vậy ...
\(\Leftrightarrow y\left(2x+1\right)=5.2x+5+\left(17-5\right)\)
\(\Leftrightarrow y\left(2x+1\right)-5\left(2x+1\right)=12\)
\(\Leftrightarrow\left(2x+1\right)\left(y-5\right)=12\) \(U\left(12\right)=\left\{\pm1,\pm2,\pm3,\pm4,\pm6,\pm12\right\}\)
\(\left\{\begin{matrix}2x+1=\left\{-3,-1,1,3\right\}\\y-5=\left\{-4,-12,12,4\right\}\end{matrix}\right.\)\(\left\{\begin{matrix}x=\left\{-2,-1,0,1\right\}\\y=\left\{1,-7,17,9\right\}\end{matrix}\right.\)
(x,y)=(-2,1);(-1,-4);(0,17);(1,9)
Ta có
2xy + y = 10x +17
2xy + y - 10x + 17 = 0
y(2x + 1) - 10x + 17 = 0
y(2x + 1) - 5(2x + 1) = 12
(2x + 1)(y - 5) = 12
Do x \(\in\) Z => 2x + 1 \(\in\) Z
y \(\in\) Z => y - 5 \(\in\) Z
=> 2x + 1; y - 5 \(\in\) Ư(12); 2x+1 là số lẻ
Ta có bảng
| x | 0 | 2 | -1 | -2 |
| 2x+1 | 1 | 3 | -1 | -3 |
| y-5 | 12 | 4 | -12 | -4 |
| y | 17 | 9 | -7 | 1 |
|
So điều kiện |
TM | TM | TM | TM |
Vậy cặp số (x;y) = (0;17) (2;9) (-1;-7) (-2;1)