60⁰ 

a.

\(\overrightarrow{u}=2\left(2;1\right)-\left(3;4\right)=\left(1;-2\right)\)

\(\overrightarrow{v}=3\left(3;4\right)-2\left(7;2\right)=\left(-5;8\right)\)

\(\overrightarrow{w}=5\left(7;2\right)+\left(2;1\right)=\left(37;11\right)\)

b.

\(\overrightarrow{x}=2\left(2;1\right)+\left(3;4\right)-\left(7;2\right)=\left(0;4\right)\)

\(\overrightarrow{z}=2\left(2;1\right)-3\left(3;4\right)+\left(7;2\right)=\left(2;-8\right)\)

c.

\(\overrightarrow{w}+\overrightarrow{a}=\overrightarrow{b}-\overrightarrow{c}\Rightarrow\overrightarrow{w}=\overrightarrow{b}-\overrightarrow{c}-\overrightarrow{a}\)

\(\Rightarrow\overrightarrow{w}=\left(3;4\right)-\left(7;2\right)-\left(2;1\right)=\left(-6;1\right)\)

a: \(\overrightarrow{CN}=\dfrac{1}{2}\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{CB}\)

\(=\dfrac{1}{2}\overrightarrow{CB}+\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{CB}\)

\(=\dfrac{1}{2}\overrightarrow{u}-\overrightarrow{v}\)

Ta có: \(\left\{{}\begin{matrix}\overrightarrow{a}=m\overrightarrow{u}+\overrightarrow{v}=\left(4m+1;m+4\right)\\\overrightarrow{b}=\overrightarrow{i}+\overrightarrow{j}=\left(1;1\right)\end{matrix}\right.\)

Yêu cầu bài toán <=> cos\(\left(\overrightarrow{a};\overrightarrow{b}\right)\)=cos45^o =\(\dfrac{\sqrt{2}}{2}\)

<=> \(\dfrac{\left(4m+1\right)+\left(m+4\right)}{\sqrt{2}\sqrt{\left(4m+1\right)^2+\left(m+4\right)^2}}=\dfrac{\sqrt{2}}{2}\)

<=> \(\dfrac{5\left(m+1\right)}{\sqrt{2}\sqrt{17m^2+16+17}}=\dfrac{\sqrt{2}}{2}\)

<=> \(5\left(m+1\right)=\sqrt{17m^2+16m+17}\)  <=>\(\left\{{}\begin{matrix}m+1\ge0\\25m^2+50m+25=17m^2+16m+17\end{matrix}\right.\)

<=> m=\(-\dfrac{1}{4}\)

- Với \(m=0\) ko thỏa mãn

- Với \(m\ne0\) hai vecto cùng phương khi:

\(\dfrac{m^2+m-2}{m}=\dfrac{4}{2}\Leftrightarrow m^2+m-2=2m\)

\(\Rightarrow m^2-3m-2=0\Rightarrow m=\dfrac{3\pm\sqrt{17}}{2}\)