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)

\( \dfrac{1}{3}\vec{a} = \dfrac{1}{3}(-3;9) = (-1;3)\)

Suy ra : A,B đúng

\( \dfrac{2}{3}\vec{a} = \dfrac{2}{3}(-3;9) = (-2;6)\)

Suy ra: D đúng

Chọn đáp án C

\(\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.cos\left(\overrightarrow{a};\overrightarrow{b}\right)\)

a/ \(\overrightarrow{a}.\overrightarrow{b}=8.\sqrt{3}.cos30^0=12\)

b/ \(\overrightarrow{a}.\overrightarrow{b}=\sqrt{2}.6.cos45^0=6\)

c/ \(\overrightarrow{a}.\overrightarrow{b}=9.10.cos60^0=45\)

d/ \(\overrightarrow{a}.\overrightarrow{b}=5.6.cos120^0=-15\)

Gọi \(M\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MA}=\left(1-x;3-y\right)\\\overrightarrow{MB}=\left(4-x;-y\right)\\\overrightarrow{MC}=\left(2-x;-5-y\right)\end{matrix}\right.\)

\(\Rightarrow\overrightarrow{MA}+\overrightarrow{MB}-3\overrightarrow{MC}=\left(x-1;y+18\right)\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\y+18=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-18\end{matrix}\right.\)

\(\Rightarrow M\left(1;-18\right)\)

\(\overrightarrow{a}\) ⊥\(\overrightarrow{b}\)

=>-1*2+2*m=0

=>2m-2=0

=>2m=2

=>m=1

=>Chọn D

\(\left|4\cdot\overrightarrow{a}-3\cdot\overrightarrow{b}\right|=\sqrt{13}\)

=>\((4\vec{a} - 3\vec{b})^2 = 13\)

=>\(16\vert{}\vec{a}\vert{}^2-24(\vec{a}\cdot\vec{b})+9\vert{}\vec{b}\vert{}^2=13\)

=>\(16\cdot1^2-24(\vec{a}\cdot\vec{b})+9\cdot^2=13\)

=>\(25 - 24(\vec{a} \cdot \vec{b}) = 13\)

=>\(24\cdot\overrightarrow{a}\cdot\overrightarrow{b}=12\)

=>\(\overrightarrow{a}\cdot\overrightarrow{b}=\frac12\)

\(\overrightarrow{c}\) ⊥\(\left(\overrightarrow{a}-\overrightarrow{b}\right)\)

=>\(\vec{c} \cdot (\vec{a} - \vec{b}) = 0\)

=>\((x\vec{a}+y\vec{b})\cdot(\vec{a}-\vec{b})=0\)

=>\(x\vert{}\vec{a}\vert{}^2-x(\vec{a}\cdot\vec{b})+y(\vec{a}\cdot\vec{b})-y\vert{}\vec{b}\vert{}^2=0\)

=>\(x - \frac{1}{2}x + \frac{1}{2}y - y = 0\)

=>x=y

\(\vec{c} = x\vec{a} + x\vec{b} = x(\vec{a} + \vec{b})\)

\(\vert{}\vec{c}\vert{}^2 = x^2(\vec{a} + \vec{b})^2 = 1\)

=>\(x^2(\vert{}\vec{a}\vert{}^2+2\vec{a}\cdot\vec{b}+\vert{}\vec{b}\vert{}^2)=1\)

=>\(x^2 \left(1 + 2 \cdot \frac{1}{2} + 1\right) = 1\)

=>\(x^2=\frac13\)

=>\(x=\pm\frac{\sqrt3}{3}\)

Khi \(x=\frac{\sqrt3}{3}\) thì \(y=x=\frac{\sqrt3}{3}\)

Khi \(x=-\frac{\sqrt3}{3}\) thì \(y=x=-\frac{\sqrt3}{3}\)