\(\frac{x}{4}-\frac{1}{y}=\frac{3}{4}\)
\(\frac{1}{y}=\frac{x-3}{4}\)
\(\left(x-3\right)\times y=4=\left(-1\right)\times\left(-4\right)=\left(-4\right)\times\left(-1\right)=4\times1=1\times4=2\times2=\left(-2\right)\times\left(-2\right)\)
Vậy \(\left(x;y\right)\in\left\{\left(2;-4\right);\left(-1;-1\right);\left(7;1\right);\left(4;4\right);\left(5;2\right);\left(1;-2\right)\right\}\)
\(\frac{x}{4}\)-\(\frac{1}{y}\)=\(\frac{3}{4}\)
\(\frac{1}{y}\)=\(\frac{x-3}{4}\)
\(\Rightarrow\)y.(x-3)=4 hay y và x-3 \(\in\)Ư(4)
Ta có bảng sau:
y | 1 | -1 | 2 | -2 | 4 | -4 |
x-1 | 4 | -4 | 2 | -2 | 1 | -1 |
x | 5 | -3 | 3 | -1 | 2 | 0 |
Vậy (x;y)\(\in\){(5;1);(-3;-1);(3;2);(-1;-2);(2;4);(0;-4)}