Cho hai điểm $A\left( {5;3;1} \right),B\left( {1;3;5} \right)$. Độ dài véc tơ $\overrightarrow {AB}$ là:

$4,2\%$

$4,26\%$

$4,3\%$

$4,5\%$

$\left[ \begin{array}{l}m = 1\\m = 2\end{array} \right.$      

$\left[ \begin{array}{l}m < 1\\m > 2\end{array} \right.$

$1 < m < 2$

Không xác định được

$\dfrac{a}{{\sqrt 3 }}.$ 

$\dfrac{{2a}}{{\sqrt 3 }}.$

$\dfrac{a}{3}.$

$\dfrac{{\sqrt{3}a}}{2}.$

$\left( {x;y} \right) = \left( { - 3 - \sqrt 3 ; - 2 + \sqrt 3 } \right)$ $\left( {x;y} \right) = \left( { - 3 + \sqrt 3 ; - 2 - \sqrt 3 } \right)$.

$\left( {x;y} \right) = \left( { - 3 - \sqrt 3 ; - 3 + \sqrt 3 } \right);$ $\left( {x;y} \right) = \left( { - 2 - \sqrt 3 ; - 2 + \sqrt 3 } \right)$.

$\left( {x;y} \right) = \left( { - 3; - 2} \right);$ $\left( {x;y} \right) = \left( {3;2} \right)$

$\left( {x;y} \right) = \left( { - 3;3} \right);$ $\left( {x;y} \right) = \left( {2;2} \right)$.

$x \in \left[ { - \dfrac{1}{3};\dfrac{1}{3}} \right].$

$x \in \left( { - \infty ; - \dfrac{1}{3}} \right) \cup \left( {\dfrac{1}{3}; + \infty } \right).$

$x \in \left( { - \infty ; - \dfrac{1}{3}} \right] \cup \left[ {\dfrac{1}{3}; + \infty } \right).$

$x \in \left( { - \dfrac{1}{3};\dfrac{1}{3}} \right).$

${(x - 2)^2} + {(y - 1)^2} = 17$

${(x - 2)^2} + {(y - 1)^2} = 2\sqrt 5$

${(x - 4)^2} + {(y - 2)^2} = 17$

${(x + 2)^2} + {(y + 1)^2} = 17$