Cho $f\left( x \right) = \sin 2x\sqrt {1 - {{\cos }^2}x}$. Nếu đặt $\sqrt {1 - {{\cos }^2}x}  = t$ thì:

$N\left( {0;2; - 1} \right)$

$N\left( {2;0;0} \right)$

$N\left( {0;2;0} \right)$        

$N\left( {0;2;1} \right)$

 $N(1;2;0)$.                           

 $M(0;0;3)$.                          

 $P(1;0;0)$.                           

$4$

$2$

$1$

$\dfrac{1}{6}$

$f\left( x \right) \geqslant  - 2,\forall x \in \left[ {1;3} \right]$ 

$f\left( 1 \right) = f\left( 3 \right) =  - 2$

$f\left( x \right) <  - 2,\forall x \in \left[ {1;3} \right]$                     

$f\left( x \right) \leqslant  - 2,\forall x \in \left[ {1;3} \right]$

$\overrightarrow {{u_1}} .\overrightarrow {{u_2}}  = 0$

$\overrightarrow {{u_1}} .\overrightarrow {{u_2}}  = \overrightarrow 0$       

$\left[ {\overrightarrow {{u_1}} ,\overrightarrow {{u_2}} } \right] = \overrightarrow 0$        

$\left[ {\overrightarrow {{u_1}} ,\overrightarrow {{u_2}} } \right] = 0$

$\dfrac{1}{2}\left| {\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right]} \right|$

$\dfrac{1}{2}\left| {\left[ {\overrightarrow {AB} ,\overrightarrow {CA} } \right]} \right|$        

$\dfrac{1}{2}\left| {\left[ {\overrightarrow {BC} ,\overrightarrow {BA} } \right]} \right|$

$\dfrac{1}{2}\left| {\left[ {\overrightarrow {AB} ,\overrightarrow {AB} } \right]} \right|$

$y = {\left( {\dfrac{1}{3}} \right)^x}$

$y = {2^x}$

$y = 3{x^3}$

$y = {\left( {\dfrac{1}{3}} \right)^{ - x}}$