Theo hệ thức Vi-ét a,b là nghiệm của phương trình:
x2-7x+12=0
\(\Delta=\left(-7\right)^2-4.1.12=1>0\)\(\Rightarrow\)phương trình có 2 nghiệm phân biệt
\(x_1=\dfrac{-\left(-7\right)+\sqrt{1}}{2}=4\)
\(x_2=\dfrac{-\left(-7\right)-\sqrt{1}}{2}=3\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=3\\b=4\end{matrix}\right.\\\left\{{}\begin{matrix}a=4\\b=3\end{matrix}\right.\end{matrix}\right.\)
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 \(\left(a;b\right)=\left(4;3\right)\)
