\(\overrightarrow{AB}.\overrightarrow{CB}+\overrightarrow{AC}.\overrightarrow{BC}=12\)

\(\Leftrightarrow\overrightarrow{BC}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)=12\)

\(\Leftrightarrow\overrightarrow{BC}.\overrightarrow{BC}=12\)

\(\Rightarrow BC^2=12\Rightarrow BC=2\sqrt{3}\)

\(\left|\overrightarrow{AB}+\overrightarrow{AC}\right|=BC=4\)

Tui cũng dag tìm câu này nè

a: A(1;3); B(-2;5); C(-4;0)

\(\overrightarrow{AB}=\left(-2-1;5-3\right)=\left(-3;2\right)\)

\(\overrightarrow{AC}=\left(-4-1;0-3\right)=\left(-5;-3\right)\)

\(\overrightarrow{BC}=\left(-4+2;0-5\right)=\left(-2;-5\right)\)

\(\overrightarrow{CB}=\left(-2+4;5-0\right)=\left(2;5\right)\)

b: \(\overrightarrow{AB}\cdot\overrightarrow{CB}=-3\cdot2+2\cdot5=-6+10=4\)

\(\overrightarrow{AC}\cdot\overrightarrow{BC}=\left(-5\right)\cdot\left(-2\right)+\left(-3\right)\cdot\left(-5\right)=10+15=25\)

c: \(\overrightarrow{AB}=\left(-3;2\right)\)

=>\(AB=\sqrt{\left(-3\right)^2+2^2}=\sqrt{13}\)

\(\overrightarrow{BC}=\left(-2;-5\right)\)

=>\(BC=\sqrt{\left(-2\right)^2+\left(-5\right)^2}=\sqrt{4+25}=\sqrt{29}\)

e: \(\overrightarrow{AB}+2\cdot\overrightarrow{CB}\) =\(\left(-3+2\cdot2;2+2\cdot5\right)\)

=(-3+4;2+10)

=(1;12)

Đáp án D

Gọi O là trung điểm của AM

BM=BC/2=a/2

\(\Leftrightarrow AM=\dfrac{a\sqrt{3}}{2}\)

\(\Leftrightarrow MO=\dfrac{a\sqrt{3}}{4}\)

Xét ΔOMB vuông tại M có 

\(BO^2=OM^2+BM^2\)

\(=a^2\cdot\dfrac{3}{16}+a^2\cdot\dfrac{1}{4}=a^2\cdot\dfrac{7}{16}\)

\(\Leftrightarrow BO=\dfrac{a\sqrt{7}}{4}\)

Xét ΔBMA có BO là đường trung tuyến

nên \(\overrightarrow{BM}+\overrightarrow{BA}=2\cdot\overrightarrow{BO}\)

\(\Leftrightarrow\left|\overrightarrow{BM}+\overrightarrow{BA}\right|=\dfrac{a\sqrt{7}}{2}\)