Có: \(\left\{{}\begin{matrix}a+b=3\\ab=1\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}b=3-a\\a\left(3-a\right)=1\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}-a^2+3a=1\\b=3-a\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}a^2-3a+1=0\\b=3-a\end{matrix}\right.\)
Cứ thế mà làm tiếp