- Với \(m=0\) ko thỏa mãn

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

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

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

\(\overrightarrow{a}=2\overrightarrow{i}-4\overrightarrow{j}\Rightarrow\overrightarrow{a}=\left(2;-4\right)\)

\(\overrightarrow{b}=-5\overrightarrow{i}+3\overrightarrow{j}\Rightarrow\overrightarrow{b}=\left(-5;3\right)\)

\(\Rightarrow\overrightarrow{u}=2\overrightarrow{a}-\overrightarrow{b}=2\left(2;-4\right)-\left(-5;3\right)=\left(9;-11\right)\)

60⁰ 

Lời giải:

a) Gọi vecto \(\overrightarrow{u}(m,n)\)

\(\left\{\begin{matrix} \overrightarrow{u}\perp \overrightarrow{a}\\ \overrightarrow{u}.\overrightarrow{b}=-4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 3m+6n=0\\ -2m+5n=-4\end{matrix}\right.\)

\(\Rightarrow m=\frac{8}{9}; n=\frac{-4}{9}\)

Vậy \(\overrightarrow{u}(\frac{8}{9}; \frac{-4}{9})\)

b) Gọi vecto \(\overrightarrow{v}(m,n)\)

\(\left\{\begin{matrix} \overrightarrow{v}\perp \overrightarrow{a}\\ |\overrightarrow{v}|=\sqrt{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 3m+6n=0\\ m^2+n^2=2\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} m=-2n\\ m^2+n^2=2\end{matrix}\right.\Rightarrow (-2n)^2+n^2=2\)

\(\Rightarrow n=\pm \sqrt{\frac{2}{5}}\)

\(\Rightarrow m=\mp 2\sqrt{\frac{2}{5}}\) (tương ứng)

Vậy..............

Câu b

\(\overrightarrow{u}+2\overrightarrow{v}-3\overrightarrow{w}+\overrightarrow{x}=\overrightarrow{0}\)

\(\Rightarrow\overrightarrow{x}=3\overrightarrow{w}-\overrightarrow{u}-2\overrightarrow{v}=3\left(-5;7\right)-\left(2;-5\right)-2\left(3;4\right)=\left(-23;18\right)\)