a) ta có : \(\overrightarrow{BA}+\overrightarrow{BC}=2\overrightarrow{BN}\) \(\Rightarrow\left|\overrightarrow{BA}+\overrightarrow{BC}\right|=2\left|\overrightarrow{BN}\right|=2BN\)
\(=2\left(AB^2-NA^2\right)=2\left(a^2-\left(\dfrac{1}{2}a\right)^2\right)=\dfrac{3}{2}a^2\)
b) \(\overrightarrow{NB}\)
c) ta có : \(\overrightarrow{NA}+\overrightarrow{MB}+\overrightarrow{PC}=\overrightarrow{NA}+\overrightarrow{AM}+\overrightarrow{PC}=\overrightarrow{NM}+\overrightarrow{PC}\)
\(=\overrightarrow{NM}+\overrightarrow{MN}=\overrightarrow{0}\left(đpcm\right)\)
d) ta có : \(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MN}+\overrightarrow{MP}+\overrightarrow{MC}=\overrightarrow{MA}+\overrightarrow{AM}+\overrightarrow{MN}+\overrightarrow{NC}+\overrightarrow{MC}\)
\(\overrightarrow{MC}+\overrightarrow{MC}=2\overrightarrow{MC}\)
\(\Rightarrow\left|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MN}+\overrightarrow{MP}+\overrightarrow{MC}\right|=2\left|\overrightarrow{MC}\right|=2MC\)
\(=2\left(AC^2-AM^2\right)=2\left(a^2-\left(\dfrac{1}{2}a\right)^2\right)=\dfrac{3}{2}a^2\)
