Lười đánh máy nên luyện chữ :))

\(\overrightarrow{BM}=\dfrac{\overrightarrow{BA}+\overrightarrow{BC}}{2}=\dfrac{\overrightarrow{BA}+\overrightarrow{BA}+\overrightarrow{AC}}{2}=-\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\)

\(\overrightarrow{AN}=\overrightarrow{AB}+\overrightarrow{BN}=\overrightarrow{AB}+\dfrac{2}{5}\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{AC}=\dfrac{3}{5}\overrightarrow{AB}+\dfrac{2}{5}\overrightarrow{AC}\)

\(\overrightarrow{NP}=\overrightarrow{NC}+\overrightarrow{CP}\)

\(=\dfrac{2}{3}\overrightarrow{BC}+\dfrac{1}{3}\overrightarrow{CA}\)

\(=-\dfrac{2}{3}\overrightarrow{CB}+\dfrac{1}{3}\overrightarrow{CA}\)

\(\overrightarrow{PM}=\overrightarrow{PA}+\overrightarrow{AM}\)

\(=\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{AB}\)

\(=\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\left(\overrightarrow{AC}+\overrightarrow{CB}\right)\)

\(=\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\)

a) Ta có:

\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)

         \(=\overrightarrow{AB}+k\overrightarrow{BC}\)

         \(=\overrightarrow{AB}+k\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\)

         \(=\left(1-k\right)\overrightarrow{AB}+k\overrightarrow{AC}\)

b) \(\overrightarrow{NP}=\overrightarrow{AP}-\overrightarrow{AN}\)

             \(=\dfrac{2}{3}\overrightarrow{AC}-\dfrac{3}{4}\overrightarrow{AB}\)

Để \(AM\perp NP\)

\(\Rightarrow\overrightarrow{AM}.\overrightarrow{NP}=\overrightarrow{0}\)

\(\Rightarrow\left[\left(1-k\right)\overrightarrow{AB}+k\overrightarrow{AC}\right]\left(-\dfrac{3}{4}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\right)=\overrightarrow{0}\)

\(\Leftrightarrow\dfrac{3\left(k-1\right)}{4}AB^2+\dfrac{2k}{3}AC^2+\dfrac{2\left(1-k\right)}{3}\overrightarrow{AB}.\overrightarrow{AC}-\dfrac{3k}{4}\overrightarrow{AB}.\overrightarrow{AC}=\overrightarrow{0}\)

\(\Leftrightarrow\dfrac{3\left(k-1\right)}{4}AB^2+\dfrac{2k}{3}AB^2+\dfrac{1-k}{3}AB^2-\dfrac{3k}{8}AB^2=0\)

\(\Leftrightarrow AB^2\left[\dfrac{3\left(k-1\right)}{4}+\dfrac{2k}{3}+\dfrac{1-k}{3}-\dfrac{3k}{8}\right]=0\)

\(\Leftrightarrow18\left(k-1\right)+16k+8\left(1-k\right)-9k=0\left(AB>0\right)\)

\(\Leftrightarrow17k=10\)

\(\Leftrightarrow k=\dfrac{10}{17}\)

\(\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AE}\)

\(\overrightarrow{AM}+\overrightarrow{AN}=2\overrightarrow{AE}\)

\(\Rightarrow\overrightarrow{AB}+\overrightarrow{AC}=\overrightarrow{AM}+\overrightarrow{AN}\)