$c$) `x/3 - 1/y = 1/3`
`⇔ 1/y = {x-1}/3`
`⇔ y(x-1) = 3`
`⇒` `y` và `x-1` `∈` `Ư(3)={±1;±3}`
Ta có bảng :
$\left[\begin{array}{ccc}y&-3&-1&1&3\\x-1&-1&-3&3&1\\x&0&-2&4&2\end{array}\right]$
Vậy `(x;y)=(0;-3);(-2;-3);(4;3);(2;1)`
$d$) `1/x + y/2 = -1/3`
`⇔ 1/x = -1/3 - y/2`
`⇔ 1/x = {-2-3y}/6`
`⇔ x(-2-3y) = 6`
`⇒` `-x;3y+2` `∈` `Ư(6)={±1;±2;±3;±6}`
Mà $3y+2$ chia $3$ dư $2$ $⇒$ $3y+2$ $∈$ `{-1;2}`
Ta có bảng
$\left[\begin{array}{ccc}3y+2&-1&2\\x&6&-3&\\y&-1&0\end{array}\right]$
Vậy `(x;y)=(6;-1);(-3;0)`
