\(a.\Leftrightarrow a\left(b+5\right)-7\left(b+5\right)+35=0\)
\(\Leftrightarrow\left(a-7\right)\left(b+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-7=0\\b+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a=7\\b=-5\end{matrix}\right.\)
Mà \(b\ge3\Rightarrow b=-5\left(loại\right)\)
Vậy a = 7 còn b bất kì thìab-7b+5a=0
\(b.\Leftrightarrow a\left(7-b\right)-2\left(7-b\right)=18-14\)
\(\Leftrightarrow\left(a-2\right)\left(7-b\right)=4\)
\(Mà\)\(4=1.4=4.1=2.2=\left(-2\right).\left(-2\right)=\left(-1\right).\left(-4\right)=\left(-4\right).\left(-1\right)\)
| a-2 | 1 | 4 | 2 | -2 | -1 | -4 |
| 7-b | 4 | 1 | 2 | -2 | -4 | -1 |
| a | 3 | 6 | 4 | 0 | 1 | -2 |
| b | 3 | 6 | 5 | 9 | 11 | 8 |
\(Vậy\)\(\left(a,b\right)=\left(3,3\right)=\left(6,6\right)=\left(0,9\right)=\left(1,11\right)=\left(-2,8\right)\)
