Ta có :
\(F\left(x\right)=ax+b\left(a\ne0\right)\)
+) \(F\left(2\right)=0\)
\(\Leftrightarrow a.2+b=0\)
\(\Leftrightarrow2a=-b\)
\(\Leftrightarrow\dfrac{a}{-1}=\dfrac{b}{2}\)
Đặt \(\dfrac{a}{-1}=\dfrac{b}{2}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=-1k\\b=2k\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{b-2014a}{a+b}=\dfrac{2k-2014k}{-1k+2k}=\dfrac{k\left(2-2014\right)}{k\left(-1+2\right)}=\dfrac{-2012}{1}=-2012\)