\(cos\left(\overrightarrow{a},\overrightarrow{b}\right)=\dfrac{1\cdot\left(-1\right)+\left(-2\right)\cdot\left(-3\right)}{\sqrt{1^2+2^2}\cdot\sqrt{1^2+3^2}}=\dfrac{5}{\sqrt{5}\cdot\sqrt{10}}=\dfrac{5}{\sqrt{50}}=\dfrac{1}{\sqrt{2}}\)

\(m\overrightarrow{a}=m\left(-1;-2\right)=\left(-m;-2m\right)\)

\(n\overrightarrow{b}=n\left(1;-3\right)=\left(n;-3n\right)\)

\(\Rightarrow m\overrightarrow{a}+n\overrightarrow{b}=\left(-m+n;-2m-3n\right)\)

\(\Rightarrow\left\{{}\begin{matrix}-m+n=2\\-2m-3n=-4\end{matrix}\right.\) \(\Rightarrow m-n=-2\) (đảo dấu pt đầu là ra, ko cần giải hẳn ra m; n)

\(\overrightarrow{NP}=\overrightarrow{NC}+\overrightarrow{CP}\)

\(=\dfrac{2}{3}\overrightarrow{BC}+\dfrac{1}{3}\overrightarrow{CA}\)

\(=-\dfrac{2}{3}\overrightarrow{CB}+\dfrac{1}{3}\overrightarrow{CA}\)

\(\overrightarrow{PM}=\overrightarrow{PA}+\overrightarrow{AM}\)

\(=\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{AB}\)

\(=\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\left(\overrightarrow{AC}+\overrightarrow{CB}\right)\)

\(=\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\)

Câu 1: 

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

\(=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)

\(=\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)

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

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

Tọa độ vecto AB là:

\(\left\{{}\begin{matrix}x=x_B-x_A=1-3=-2\\y=y_B-y_A=-3-2=-5\end{matrix}\right.\)

Vậy: \(\overrightarrow{AB}=\left(-2;-5\right)\)

Tọa độ vecto AC là:

\(\left\{{}\begin{matrix}x=x_C-x_A=1-3=-2\\y=y_C-y_A=4-2=2\end{matrix}\right.\)

Vậy: \(\overrightarrow{AC}=\left(-2;2\right)\)

=>\(\overrightarrow{CA}=\left(2;-2\right)\)

Tọa độ vecto BC là:

\(\left\{{}\begin{matrix}x=x_C-x_B=1-1=0\\y=y_C-y_B=4-\left(-3\right)=7\end{matrix}\right.\)

Vậy: \(\overrightarrow{BC}=\left(0;7\right)\)

b: \(\overrightarrow{AB}=\left(-2;-5\right);\overrightarrow{AC}=\left(-2;2\right);\overrightarrow{BC}=\left(0;7\right)\)

\(AB=\sqrt{\left(-2\right)^2+\left(-5\right)^2}=\sqrt{29}\)

\(AC=\sqrt{\left(-2\right)^2+2^2}=2\sqrt{2}\)

\(BC=\sqrt{0^2+7^2}=7\)

Chu vi tam giác ABC là:

\(AB+AC+BC=2\sqrt{2}+\sqrt{29}+7\)