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a/ \(\overrightarrow{AN}+\overrightarrow{BP}+\overrightarrow{CM}=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)+\frac{1}{2}\left(\overrightarrow{BC}+\overrightarrow{BA}\right)+\frac{1}{2}\left(\overrightarrow{CA}+\overrightarrow{CB}\right)\)
\(=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{BA}\right)+\frac{1}{2}\left(\overrightarrow{AC}+\overrightarrow{CA}\right)+\frac{1}{2}\left(\overrightarrow{BC}+\overrightarrow{CB}\right)=\overrightarrow{0}\)
b/
Do MN là đường trung bình tam giác ABC \(\Rightarrow\overrightarrow{MN}=\frac{1}{2}\overrightarrow{AC}\)
\(\overrightarrow{AN}=\overrightarrow{AM}+\overrightarrow{MN}=\overrightarrow{AM}+\frac{1}{2}\overrightarrow{AC}=\overrightarrow{AM}+\overrightarrow{AP}\)
c/
\(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}+\frac{1}{2}\overrightarrow{CA}=\frac{1}{2}\overrightarrow{AC}+\frac{1}{2}\overrightarrow{CA}=\overrightarrow{0}\)
a: \(\overrightarrow{AM}+\overrightarrow{BN}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{BC}=\dfrac{1}{2}\overrightarrow{AC}\)
b: \(=\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{BA}\)
\(=\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{BA}\)
c: \(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}\)
\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{BC}+\dfrac{1}{2}\overrightarrow{CA}\)
\(=\dfrac{1}{2}\left(\overrightarrow{AC}+\overrightarrow{CA}\right)=\overrightarrow{0}\)