Theo các xác định điểm M, N ta có:
\(\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB};\overrightarrow{AN}=\dfrac{2}{3}\overrightarrow{AC}.\)
Theo tính chất trung điểm của MN ta có:
\(\overrightarrow{AK}=\dfrac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\right)\)
\(=\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\).

Chọn B.

Ta có 

mà 

E cần gấp achij nào giúp e cho mai e nộp

a) \(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AN}=\dfrac{-1}{2}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)

b) CG.CAN??

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 \(AM\perp NP\)

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

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

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

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

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

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

\(\Leftrightarrow17k=10\)

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

a)các vectow cùng phương với AM LÀ: MA ;MB;BM;BA;AB;PN;NP

b)các vectow cùng hướng  MN là:BP;PC;BC

c)các vectow ngược hướng với BC là:CP;CP;NM

Xét ΔBAD có BM là đường trung tuyến

nên \(\overrightarrow{BM}=\dfrac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\)

\(=\dfrac{1}{2}\left(\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{BC}\right)\)

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

\(=\dfrac{1}{2}\left(\dfrac{5}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{AC}\right)\)

\(=\dfrac{1}{6}\left(5\overrightarrow{BA}+2\overrightarrow{AC}\right)\)

\(=\dfrac{5}{6}\left(\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{AC}\right)\)

\(\overrightarrow{BN}=\overrightarrow{BA}+\overrightarrow{AN}\)

\(=\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{BC}\)

=>\(\overrightarrow{BM}=\dfrac{5}{6}\cdot\overrightarrow{BN}\)

=>B,M,N thẳng hàng