a: \(\overrightarrow{x}+\overrightarrow{BC}=\overrightarrow{AC}+\overrightarrow{BA}\)

=>\(\overrightarrow{x}+\overrightarrow{BC}=\overrightarrow{BC}\)

=>\(\overrightarrow{x}=\overrightarrow{0}\)

b: \(\overrightarrow{CA}-\overrightarrow{x}-\overrightarrow{CB}=\overrightarrow{AB}\)

=>\(\overrightarrow{BC}+\overrightarrow{CA}-\overrightarrow{x}=\overrightarrow{AB}\)

=>\(\overrightarrow{BA}-\overrightarrow{x}=\overrightarrow{AB}\)

=>\(\overrightarrow{x}=\overrightarrow{BA}-\overrightarrow{AB}=2\cdot\overrightarrow{BA}\)

có ai biết cách làm thì giúp mk với mai mk cần lắm rồi

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

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

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

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

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

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

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

b: \(\left|\overrightarrow{BA}+\overrightarrow{BC}\right|=\left|\dfrac{\overrightarrow{AC}}{2}\right|=\dfrac{5}{2}a\)

\(\overrightarrow{MC}=3\cdot\overrightarrow{MA}\)

=>MC=3MA và A nằm giữa M và C

MA+AC=MC

=>AC=MC-MA=3MA-MA=2MA

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

\(=\overrightarrow{BA}+2\cdot\overrightarrow{CA}=\overrightarrow{BA}+2\cdot\overrightarrow{CB}+2\cdot\overrightarrow{BA}=3\cdot\overrightarrow{BA}-2\cdot\overrightarrow{BC}\)