Ta có : \(\frac{1}{x}-\frac{y}{6}=\frac{1}{3}\Leftrightarrow\frac{1}{x}=\frac{1}{3}+\frac{y}{6}\Leftrightarrow\frac{1}{x}=\frac{2+y}{6}\)
\(\Leftrightarrow\left(2+y\right)x=6\Leftrightarrow2+y;x\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
x | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 |
2 + y | 6 | -6 | 3 | -3 | 2 | -2 | 1 | -1 |
y | 4 | -8 | 1 | -5 | 0 | -4 | -1 | -3 |
\(\frac{1}{x}-\frac{y}{6}=\frac{1}{3}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{3}+\frac{y}{6}\)
\(\Rightarrow\frac{1}{x}=\frac{2}{6}+\frac{y}{6}\)
\(\Rightarrow\frac{1}{x}=\frac{2+y}{6}\)
\(\Rightarrow x\left(2+y\right)=6\)
Ta có bảng sau :
x | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 |
2+y | 6 | -6 | 3 | -3 | 2 | -2 | 1 | -1 |
y | 4 | -8 | 1 | -5 | 0 | -4 | -1 | -3 |
Vậy ( x ; y ) = { ( 1 ; 4 ) , ( -1 ; -8 ) , ( 2 ; 1 ) , ( -2 ; -5 ) , ( 3 ; 0 ) , ( -3 ; -4 ) , ( 6 ; -1 ) , ( -6 ; -3 ) }
ta có: \(\frac{1}{x}-\frac{y}{6}\)=\(\frac{1}{3}\)<=>\(\frac{1}{x}\)=\(\frac{1}{3}\)+\(\frac{y}{6}\)
<=>\(\frac{1}{x}\)=\(\frac{2+y}{6}\)<=>x(2+y)=6
Mà x, y nguyên => x và y+2 ∈Ư(6)={±1;±2;±3;±6}
thay vào ta tìm được các cặp x,y.