Ta có :
\(\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}\) (\(M\) là trung điểm \(AB\))
\(\overrightarrow{AN}=\dfrac{2}{3}\overrightarrow{AC}\left(AN=2NC\right)\)
\(I\) là trung điểm \(MN\)
\(\Rightarrow\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\right)=\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\left(1\right)\)
\(\overrightarrow{BP}=k\overrightarrow{BC}\left(k=\dfrac{a}{b}\right)\)
\(\Leftrightarrow\overrightarrow{AP}-\overrightarrow{AB}=k\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\)
\(\Leftrightarrow\overrightarrow{AP}=\left(1-k\right)\overrightarrow{AB}+k\overrightarrow{AC}\)
mà \(\overrightarrow{AI}=x\overrightarrow{AP}\left(I\in AP\right)\left(0< x< 1\right)\)
\(\left(1\right)\Rightarrow\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=x\left[\left(1-k\right)\overrightarrow{AB}+k\overrightarrow{AC}\right]\)
\(\Leftrightarrow\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}==x\left(1-k\right)\overrightarrow{AB}+kx\overrightarrow{AC}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(1-k\right)=\dfrac{1}{4}\\kx=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}k=\dfrac{4}{7}\left(tm\right)\\x=\dfrac{7}{12}\left(tm\right)\end{matrix}\right.\)
\(\Rightarrow k=\dfrac{a}{b}=\dfrac{4}{7}\Rightarrow\left\{{}\begin{matrix}a=4\\b=7\end{matrix}\right.\) \(\left(a;b=1\right)\)
\(\Rightarrow a+b=4+7=11\)