\(\frac{a}{7}-\frac{1}{2}=\frac{1}{b+1}\\ \Rightarrow\frac{2a-7}{14}=\frac{1}{b+1}\\ \Rightarrow\left(2a-7\right)\left(b+1\right)=14\)
Ta có bảng sau:
\(2a-7\) | 1 | 14 | -1 | -14 | 2 | 7 | -2 | -7 |
\(b+1\) | 14 | 1 | -14 | -1 | 7 | 2 | -7 | -2 |
a | 4 | \(\frac{21}{2}\) | 3 | \(\frac{-7}{2}\) | \(\frac{9}{2}\) | 7 | \(\frac{5}{2}\) | 0 |
b | 13 | 0 | -15 | -2 | 6 | 1 | -8 | -3 |
Xét ĐK | t/m | ktm | t/m | ktm | ktm | t/m | ktm | t/m |
Vậy \(\left(a;b\right)\in\left\{\left(4;13\right);\left(3;-15\right);\left(7;1\right);\left(0;-3\right)\right\}\)