Ta có:
\(\left\{{}\begin{matrix}a+b=7\\ab=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=7-b\\\left(7-b\right)=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=7-b\\b^2-7b+12=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=7-b\\\left(b-4\right)\left(b-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a=3\\a=4\end{matrix}\right.\\\left[{}\begin{matrix}b=4\\b=3\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(a;b\right)=\left(3;4\right)\) hoặc \(\left(a;b\right)=\left(4;3\right)\)