Ta có:
\(\frac{x}{4}-\frac{1}{y}=\frac{1}{2}\)
\(\frac{xy}{4y}-\frac{4}{4y}=\frac{1}{2}\)
\(\frac{xy-4}{4y}=\frac{1}{2}\)
\(2.\left(xy-4\right)=4y\)
\(2xy-8-4y=0\)
\(2xy-2-4-4y=0\)
\(2.\left(xy+1\right)-4.\left(y+1\right)=0\)
\(2.\left(xy+1\right)-2.2.\left(y+1\right)=0\)
\(2.\left[\left(xy+1\right)-2.\left(y+1\right)\right]=0\)
\(xy+1-2y-2=0\)
\(y.\left(x-2\right)=1\)
Ta có:1=1.1=(-1).(-1)
Do đó ta có bảng sau:
| y | 1 | -1 |
| x-2 | 1 | -1 |
| x | 3 | 1 |
Vậy cặp (x;y) TM là:(3;1)(1;-1)
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