Làm giúp với ạ. Giúp ạ.

$\overrightarrow{MB} = 3\overrightarrow{MC} \Rightarrow \overrightarrow{BM} = \dfrac{3}{2}\overrightarrow{BC}$

$\overrightarrow{NA} = 3\overrightarrow{CN} \Rightarrow \overrightarrow{AN} = \dfrac{3}{4}\overrightarrow{AC}$

$\overrightarrow{PA} + \overrightarrow{PB} = \overrightarrow{0} \Rightarrow \overrightarrow{PA} = -\dfrac{1}{2}\overrightarrow{AB}$

a) Ta có:

$+) \quad \overrightarrow{PN} = \overrightarrow{PA} + \overrightarrow{AN}$

$= – \dfrac{1}{2}\overrightarrow{AB} + \dfrac{3}{4}\overrightarrow{AC}$

$+) \quad \overrightarrow{PM} = \overrightarrow{PA} + \overrightarrow{AM}$

$=- \dfrac{1}{2}\overrightarrow{AB} + \overrightarrow{AB} + \overrightarrow{BM}$

$= \dfrac{1}{2}\overrightarrow{AB} + \dfrac{3}{2}\overrightarrow{BC}$

$= \dfrac{1}{2}\overrightarrow{AB} + \dfrac{3}{2}(\overrightarrow{BA} + \overrightarrow{AC})$

$= \dfrac{1}{2}\overrightarrow{AB} – \dfrac{3}{2}(\overrightarrow{AB} + \dfrac{3}{2}\overrightarrow{AC})$

$= -\overrightarrow{AB} + \dfrac{3}{2}\overrightarrow{AC}$

b) Ta có: $\overrightarrow{PN} = – \dfrac{1}{2}\overrightarrow{AB} + \dfrac{3}{4}\overrightarrow{AC}$

$\overrightarrow{PM} = -\overrightarrow{AB} + \dfrac{3}{2}\overrightarrow{AC} = 2\left(- \dfrac{1}{2}\overrightarrow{AB} + \dfrac{3}{4}\overrightarrow{AC}\right) = 2\overrightarrow{PN}$

$\Rightarrow M,N,P$ thẳng hàng

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Áp dụng định lý $Menelaus$ ta có:

$\dfrac{PA}{PB}\cdot\dfrac{MB}{MC}\cdot\dfrac{NC}{NA}$

$= 1\cdot3\cdot\dfrac{1}{3} = 1$

$\Rightarrow M,N,P$ thẳng hàng