Cho hai véc tơ $\overrightarrow u  = \left( {a;0;1} \right),\overrightarrow v  = \left( { - 2;0;c} \right)$. Biết $\overrightarrow u  = \overrightarrow v$, khi đó:

$- \overrightarrow a$ hoặc $- \overrightarrow b$

$\left[ {\overrightarrow a ,\overrightarrow b } \right]$    

$\overrightarrow a  - \overrightarrow b$

$\overrightarrow a  + \overrightarrow b$

${{e}^{-2}}$                          

${{e}^{3}}-e$             

$e-{{e}^{3}}$                         

$\left| {{z_1}} \right| + \left| {{z_2}} \right| = 4$.

$\left| {{z_1}} \right| + \left| {{z_2}} \right| = 2\sqrt 5$.

$\left| {{z_1}} \right| + \left| {{z_2}} \right| = 10$.

$\left| {{z_1}} \right| + \left| {{z_2}} \right| = \sqrt 5$.

$\int {0dx}  = C$       

$\int {dx}  = C$         

$\int {dx}  = 0$          

$\int {0dx}  = x + C$

${(x + 1)^2} + {(y + 2)^2} + {(z - 3)^2} = 53.$

${(x + 1)^2} + {(y + 2)^2} + {(z + 3)^2} = 53.$

${(x - 1)^2} + {(y - 2)^2} + {(z - 3)^2} = 53.$

${(x - 1)^2} + {(y - 2)^2} + {(z + 3)^2} = 53.$

 $-3\sin x+\dfrac{1}{x}+C$.

 $3\sin x-\dfrac{1}{x}+C$. 

$3\cos x+\dfrac{1}{x}+C$.