Cho tam giác ABC. Gọi M, N, P là các điểm được xác...

A

- Hướng dẫn giải

* Ta có:

$\begin{array}{l}\overrightarrow{MC}=3\overrightarrow{MB}\iff \overrightarrow{MC}=3(\overrightarrow{MC}+\text{}\overrightarrow{CB})\iff -2\overrightarrow{MC}=3\overrightarrow{CB}\\ \iff \overrightarrow{CB}=\frac{-2}{3}\overrightarrow{MC}\iff \overrightarrow{BC}=\frac{2}{3}\overrightarrow{MC}\end{array}$  

* $\begin{array}{l}\overrightarrow{NA}=-2\overrightarrow{NB}\iff \overrightarrow{NA\text{}}=-2(\overrightarrow{NA}+\text{}\overrightarrow{AB})\\ \iff 3\overrightarrow{NA}=-2A\overrightarrow{B}\iff \overrightarrow{AB}=\frac{-3}{2}\overrightarrow{NA}\end{array}$  

 Do đó,

$\begin{array}{l}\overrightarrow{AP}=x.\overrightarrow{AC}=x.\left(\overrightarrow{AB}+\text{}\overrightarrow{BC}\right)\\ =x.\left(\frac{-3}{2}\overrightarrow{NA}+\text{}\frac{2}{3}\overrightarrow{MC}\right)=\frac{-3}{2}x\overrightarrow{NA}+\text{\hspace{0.17em}\hspace{0.17em}}\frac{2}{3}x\overrightarrow{MC}\\ =\frac{3}{2}x\overrightarrow{AN}+\text{\hspace{0.17em}\hspace{0.17em}}\frac{2}{3}x(\overrightarrow{AC}-\overrightarrow{AM\text{}})\\ =\frac{3}{2}x\overrightarrow{AN}+\text{\hspace{0.17em}\hspace{0.17em}}\frac{2}{3}x.(\frac{1}{x}.\overrightarrow{AP}-\overrightarrow{AM\text{}})\\ =\frac{3}{2}x\overrightarrow{AN}+\text{\hspace{0.17em}}\frac{2}{3}.\overrightarrow{AP}-\frac{2}{3}x\overrightarrow{AM\text{}}\end{array}$

 $\begin{array}{l}\iff \frac{1}{3}\overrightarrow{AP}=\frac{3}{2}x\overrightarrow{AN}-\frac{2}{3}x\overrightarrow{AM\text{}}\\ \iff \overrightarrow{AP}=\frac{9}{2}x\overrightarrow{AN}-2x\overrightarrow{AM\text{}}\end{array}$

Để ba điểm M; N;  P thẳng hàng thì

$\frac{9}{2}x-2x=1\iff \frac{5}{2}x=1\iff x=\frac{2}{5}$

Đáp án A