Tìm a,b thuộc Z. a) b/5 + 1/10 = 1/a b) a/4 - 1/2 = 3/b

`a)`

`b/5 + 1/10 = 1/a`

`⇔(2b)/10+1/10 = 1/a`

`⇔(2b+1)/10 = 1/a`

`⇔(2b+1)/10 = 1/a`
`<=>(2b+1)a=10` 
Do `a;b in ZZ` nên `2b+1;a in ZZ`
`=> 2b+1 ; a in Ư_{(10)} = { +-1;+-2;+-5;+-10}`

Do `2b + 1` là số lẻ `AAb in ZZ` nên `2b+ 1 in { +-1 ; +-5 }`

\begin{array}{|c|c|c|}\hline \text{2b+1}&\text{1}&\text{-1}&\text{5}&\text{-5}\\\hline \text{2b}&\text{0}&\text{-2}&\text{4}&\text{-6}\\\hline \text{b}&\text{0}&\text{-1}&\text{2}&\text{-3}\\\hline \text{a}&\text{5}&\text{-5}&\text{1}&\text{-1}\\\hline\end{array}
Vậy `(a;b) in {(5;0);(-5;-1);(1;2);(-1;-3)}` 

`b)` 

`a/4 - 1/2 = 3/b`

`<=> a/4 - 2/4 = 3/b`

`<=> ( a - 2 )/4 = 3/b` 

Do `a;b in ZZ` nên `a-2;b in ZZ`
`=> a-2;b in Ư_{(12)} = { +-1;+-2;+-3;+-4;+-6;+-12}`
Lập bảng `:`
\begin{array}{|c|c|c|}\hline \text{a-2}&\text{1}&\text{-1}&\text{2}&\text{-2}&\text{3}&\text{-3}&\text{4}&\text{-4}&\text{6}&\text{-6}&\text{12}&\text{-12}\\\hline \text{a}&\text{3}&\text{1}&\text{4}&\text{0}&\text{5}&\text{-1}&\text{6}&\text{-2}&\text{8}&\text{-4}&\text{14}&\text{-10}\\\hline \text{b}&\text{12}&\text{-12}&\text{6}&\text{-6}&\text{4}&\text{-4}&\text{3}&\text{-3}&\text{2}&\text{-2}&\text{1}&\text{-1}\\\hline\end{array}
Vậy `(a;b) in {(3;12);(1;-12);(4;6);(0;-6);(5;4);(-1;4);(6;3);(-2;-3);(8;2);(-4;-2);(14;1);(-10;-1)}`