Ta có:
\(5y-3x=2xy-11\)
\(\Leftrightarrow2xy-11-5y+3x=0\)
\(\Leftrightarrow2\cdot\left(2xy-11-5y+3x\right)=0\cdot2\)
\(\Leftrightarrow4xy-22-10y+6x=0\)
\(\Leftrightarrow4xy+6x-10y-15-7=0\)
\(\Leftrightarrow\left(4xy+6x\right)-\left(10y+15\right)=7\)
\(\Leftrightarrow2x\left(2y+3\right)-5\left(2y+3\right)=7\\ \Leftrightarrow\left(2x-5\right)\cdot\left(2y+3\right)=7\)
Ta có bảng sau:
| 2x-5 | 1 | 7 | -1 | -7 |
| 2y+3 | 7 | 1 | -7 | -1 |
| x | 3 | 6 | 2 | -1 |
| y | 2 | -1 | -5 | -2 |
Vậy có các cặp số (x;y) là : (3;2), (6;-1), (2;-5),(-1;-2)
Ta có:
5y−3x=2xy−115y−3x=2xy−11
⇔2xy−11−5y+3x=0⇔2xy−11−5y+3x=0
⇔2⋅(2xy−11−5y+3x)=0⋅2⇔2⋅(2xy−11−5y+3x)=0⋅2
⇔4xy−22−10y+6x=0⇔4xy−22−10y+6x=0
⇔4xy+6x−10y−15−7=0⇔4xy+6x−10y−15−7=0
⇔(4xy+6x)−(10y+15)=7⇔(4xy+6x)−(10y+15)=7
⇔2x(2y+3)−5(2y+3)=7⇔(2x−5)⋅(2y+3)=7⇔2x(2y+3)−5(2y+3)=7⇔(2x−5)⋅(2y+3)=7
Ta có bảng sau:
| 2x-5 | 1 | 7 | -1 | -7 |
| 2y+3 | 7 | 1 | -7 | -1 |
| x | 3 | 6 | 2 | -1 |
| y | 2 | -1 | -5 | -2 |
Vậy có các cặp số (x;y) là : (3;2), (6;-1), (2;-5),(-1;-2)
