Giả sử `\vec{c}=m\vec{a}+n\vec{b}`

`<=>(3;-4)=m(2;0)+n(0;-3)`

`<=>(3;-4)=(2m;-3n)`

`<=>{(m=3/2),(n=4/3):}`

   `=>\vec{c}=3/2\vec{a}+4/3\vec{b}`

(1); vecto u=2*vecto a-vecto b

=>\(\left\{{}\begin{matrix}x=2\cdot1-0=2\\y=2\cdot\left(-4\right)-2=-10\end{matrix}\right.\)

(2): vecto u=-2*vecto a+vecto b

=>\(\left\{{}\begin{matrix}x=-2\cdot\left(-7\right)+4=18\\y=-2\cdot3+1=-5\end{matrix}\right.\)

(3): vecto a=2*vecto u-5*vecto v

\(\Leftrightarrow\left\{{}\begin{matrix}a=2\cdot\left(-5\right)-5\cdot0=-10\\b=2\cdot4-5\cdot\left(-3\right)=15+8=23\end{matrix}\right.\)

(4): vecto OM=(x;y)

2 vecto OA-5 vecto OB=(-18;37)

=>x=-18; y=37

=>x+y=19

Chọn B

a) Vì \(\overrightarrow v  = \left( {0; - 7} \right)\)nên \(\overrightarrow v  = 0\overrightarrow i  + \left( { - 7} \right)\overrightarrow j  =  - 7\overrightarrow j \)

b) Vì B có tọa  độ là (-1; 0) nên \(\overrightarrow {OB}  = \left( { - 1;{\rm{ }}0} \right)\). Do đó: \(\overrightarrow {OB}  = \left( { - 1} \right)\overrightarrow i  + 0\overrightarrow j  =  - \overrightarrow i \)

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

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

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

\(\overrightarrow{v}=3\overrightarrow{i}-m\overrightarrow{j}\Rightarrow\overrightarrow{v}=\left(3;-m\right)\)

Để \(\overrightarrow{u};\overrightarrow{v}\) cùng phương:

\(\Leftrightarrow\frac{3}{-2}=\frac{-m}{1}\Rightarrow m=\frac{3}{2}\)

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

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=\(-\dfrac{1}{4}\)

Còn 2 ở mẫu kia thì đi đâu r ạ