a;\(\overrightarrow{AB}+2\overrightarrow{AC}\)

\(=\overrightarrow{AM}+\overrightarrow{MB}+2\overrightarrow{AM}+2\overrightarrow{MC}\)

\(=3\overrightarrow{AM}\)

b: \(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\)

\(=\overrightarrow{MG}+\overrightarrow{GA}+\overrightarrow{MG}+\overrightarrow{GB}+\overrightarrow{MG}+\overrightarrow{GC}\)

=3vecto MG

Lời giải:

Theo đề ta có: $\overrightarrow{BM}=2\overrightarrow{MC}=-2\overrightarrow{CM}$

$\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}(1)$

$=\overrightarrow{AB}-2\overrightarrow{CM}$

$\overrightarrow{AM}=\overrightarrow{AC}+\overrightarrow{CM}$

$\Rightarrow 2\overrightarrow{AM}=2\overrightarrow{AC}+2\overrightarrow{CM}(2)$

Lấy $(1)+(2)\Rightarrow 3\overrightarrow{AM}=\overrightarrow{AB}+2\overrightarrow{AC}$

$\Rightarrow \overrightarrow{AM}=\frac{1}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}$

Hình vẽ:

Hình bạn tự vẽ :

AM=AB+BM

=AB+2/3BC

=AB +2/3(BA+AC)

=AB-2/3AB+2/3C

= 1/3 AB + 2/3AC

Do G là trọng tâm ABC \(\Rightarrow\overrightarrow{BG}=\dfrac{1}{3}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}\)

I đối xứng B qua G \(\Rightarrow\) \(\overrightarrow{BI}=2\overrightarrow{BG}=\dfrac{2}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{BC}=\dfrac{2}{3}\overrightarrow{BA}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)

\(\Rightarrow\overrightarrow{BI}=\dfrac{4}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{AC}=-\dfrac{4}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)

\(\Rightarrow\overrightarrow{CI}=\overrightarrow{CB}+\overrightarrow{BI}=\overrightarrow{CA}+\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)

\(\Rightarrow\overrightarrow{CI}=-\dfrac{1}{3}\overrightarrow{AB}-\dfrac{1}{3}\overrightarrow{AC}\)

Do trắc nghiệm nên ta chỉ cần xét trường hợp đặc biệt nhất: đường thẳng này đi qua B, khi đó M trùng B và N là trung điểm AC

\(\Rightarrow\overrightarrow{AM}.\overrightarrow{AN}=\dfrac{1}{2}\overrightarrow{AB}.\overrightarrow{AC}\)

Đồng thời do \(\overrightarrow{MB}=\overrightarrow{0}\) và \(\overrightarrow{NC}=\overrightarrow{AN}=\dfrac{1}{2}\overrightarrow{AC}\) nên đáp án D đúng

\(\overrightarrow{MB}=-2\overrightarrow{MC}\Leftrightarrow\overrightarrow{MB}=-2\left(\overrightarrow{MB}+\overrightarrow{BC}\right)\)

\(\Rightarrow3\overrightarrow{MB}=-2\overrightarrow{BC}\Rightarrow\overrightarrow{BM}=\frac{2}{3}\overrightarrow{BC}=\frac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=-\frac{2}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}\)

\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}-\frac{2}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}\)

\(\Rightarrow\left\{{}\begin{matrix}m=\frac{1}{3}\\n=\frac{2}{3}\end{matrix}\right.\) \(\Rightarrow mn=\frac{2}{9}\)