\(\dfrac{x}{4}-\dfrac{1}{y}=\dfrac{1}{2}\left(y\ne0,x,y\in Z\right)\)
\(=>\dfrac{xy-4}{4y}=\dfrac{2y}{4y}=>xy-4=2y\)
\(< =>\)\(xy-2y=4\)\(< =>y\left(x-2\right)=4\)
\(=2.2=1.4=4.1=\left(-1\right)\left(-4\right)=\left(-4\right)\left(-1\right)\)
*với \(\left\{{}\begin{matrix}x-2=2\\y=2\end{matrix}\right.=>\left\{{}\begin{matrix}x=4\left(TM\right)\\y=2\left(TM\right)\end{matrix}\right.\)
*với \(\left\{{}\begin{matrix}x-2=1\\y=4\end{matrix}\right.=>\left\{{}\begin{matrix}x=3\left(TM\right)\\y=4\left(TM\right)\end{matrix}\right.\)
*vơi \(\left\{{}\begin{matrix}x-2=4\\y=1\end{matrix}\right.=>\left\{{}\begin{matrix}x=6\left(TM\right)\\y=1\left(TM\right)\end{matrix}\right.\)
*với \(\left\{{}\begin{matrix}x-2=-1\\y=-4\end{matrix}\right.=>\left\{{}\begin{matrix}x=1\left(TM\right)\\y=-4\left(TM\right)\end{matrix}\right.\)
*với \(\left\{{}\begin{matrix}x-2=-4\\y=-1\end{matrix}\right.=>\left\{{}\begin{matrix}x=-2\left(TM\right)\\y=-1\left(TM\right)\end{matrix}\right.\)
Vậy....
Giải:
\(\dfrac{x}{4}-\dfrac{1}{y}=\dfrac{1}{2}\)
\(\dfrac{1}{y}=\dfrac{x}{4}-\dfrac{1}{2}\)
\(\dfrac{1}{y}=\dfrac{x-2}{4}\)
\(\Rightarrow y.\left(x-2\right)=1.4\)
\(\Rightarrow y.\left(x-2\right)=4\)
\(\Rightarrow y\) và \(\left(x-2\right)\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Ta có bảng giá trị:
| x-2 | -4 | -2 | -1 | 1 | 2 | 4 |
| y | -1 | -2 | -4 | 4 | 2 | 1 |
| x | -2 | 0 | 1 | 3 | 4 | 6 |
Vậy \(\left(x;y\right)=\left\{\left(-2;-1\right);\left(0;-2\right);\left(1;-4\right);\left(3;4\right);\left(4;2\right);\left(6;1\right)\right\}\)
