Trắc nghiệm Hệ trục tọa độ có đáp án (Nhận biết)

$\left(i;j\right)$

$\left(\overrightarrow{i};\overrightarrow{j}\right)$

$\left(\overrightarrow{m};\overrightarrow{n}\right)$

$\left(m;n\right)$

$\overrightarrow{u}=-\overrightarrow{i}$

$\overrightarrow{u}=-\overrightarrow{j}$

$\overrightarrow{u}=\overrightarrow{i}$

$\overrightarrow{u}=\overrightarrow{j}$

$\overrightarrow{OM}=\left(-3;-1\right)$

$\overrightarrow{OM}=\left(3;1\right)$

$\overrightarrow{MO}=\left(-3;1\right)$

$\overrightarrow{OM}=\left(-3;1\right)$

$\overrightarrow{OM}=-2\overrightarrow{i}+4\overrightarrow{j}$

$\overrightarrow{OM}=2\overrightarrow{i}-4\overrightarrow{j}$

$\overrightarrow{OM}=2\overrightarrow{j}-4\overrightarrow{i}$

$\overrightarrow{OM}=-2\overrightarrow{i}-4\overrightarrow{j}$

$I\left(\frac{x_A-x_B}{2};\frac{y_A-y_B}{2}\right)$

$I\left(\frac{x_A+x_B}{2};\frac{y_A+y_B}{2}\right)$

$I\left(\frac{x_A+x_B}{3};\frac{y_A+y_B}{3}\right)$

$I\left(\frac{x_A+y_A}{2};\frac{x_B+y_B}{2}\right)$

$G\left(\frac{x_A-x_B+x_C}{3};\frac{y_A+y_B+y_C}{3}\right)$

$G\left(\frac{x_A+x_B+x_C}{3};\frac{y_A+y_B+y_C}{2}\right)$

$G\left(\frac{x_A+x_B+x_C}{3};\frac{y_A+y_B+y_C}{3}\right)$

$G\left(\frac{x_A+x_B+x_C}{2};\frac{y_A+y_B+y_C}{3}\right)$

$\left\{\begin{array}{l}{u}_1=u_2\\ v_1=v_2\end{array}\right.$

$\left\{\begin{array}{l}{u}_1=-v_1\\ u_2=-v_2\end{array}\right.$

$\left\{\begin{array}{l}{u}_1=v_1\\ u_2=v_2\end{array}\right.$

$\left\{\begin{array}{l}{u}_1=v_2\\ u_2=v_1\end{array}\right.$

$\overrightarrow{AB}=\left(y_A-x_A;y_B-x_B\right)$

$\overrightarrow{AB}=\left(x_A+x_B;y_A+y_B\right)$

$\overrightarrow{AB}=\left(x_A-x_B;y_A-y_B\right)$

$\overrightarrow{AB}=\left(x_B-x_A;y_B-y_A\right)$

(2;4)

(5;6)

(15;10)

(50;6)

Hai vectơ $\stackrel{\rightharpoonup }{u}=(2;-1)$ và $\overrightarrow{v}=(-1;2)$ đối nhau

Hai vectơ $\stackrel{\rightharpoonup }{u}=(2;-1)$ và $\overrightarrow{v}=(-2;-1)$ đối nhau

Hai vectơ $\stackrel{\rightharpoonup }{u}=(2;-1)$ và $\overrightarrow{v}=(-2;1)$ đối nhau

Hai vectơ $\stackrel{\rightharpoonup }{u}=(2;-1)$ và $\overrightarrow{v}=(2;1)$ đối nhau

$\overrightarrow{a}=(-5;0),\overrightarrow{b}=(-4;0)$ cùng hướng

$\overrightarrow{c}=(3;-7)làvectơđốicủad=(-7;3)$

$\overrightarrow{u}=(-4;2),\overrightarrow{v}=(-8;3)cùngphương$

$\overrightarrow{a}=(6;3),\overrightarrow{b}=(2;1)$ngược hướng

(-2;-4)

(-5;-6)

(5;6)

(-15;-10)

$\overrightarrow{OM}=\left(-3;-1\right)$

$\overrightarrow{OM}=\left(3;1\right)$

$\overrightarrow{MO}=\left(-3;1\right)$

$\overrightarrow{OM}=\left(-3;1\right)$

$\overrightarrow{u}=-\overrightarrow{i}$

$\overrightarrow{u}=-\overrightarrow{j}$

$\overrightarrow{u}=\overrightarrow{i}$

$\overrightarrow{u}=\overrightarrow{j}$

$\overrightarrow{OM}=-2\overrightarrow{i}+4\overrightarrow{j}$

$\overrightarrow{OM}=2\overrightarrow{i}-4\overrightarrow{j}$

$\overrightarrow{OM}=2\overrightarrow{j}-4\overrightarrow{i}$

$\overrightarrow{OM}=-2\overrightarrow{i}-4\overrightarrow{j}$