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

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

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

\(N\left( {0;2;0} \right)\)        

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

 \(N(1;2;0)\).                           

 \(M(0;0;3)\).                          

 \(P(1;0;0)\).                           

\(4\)

\(2\)

\(1\)

\(\dfrac{1}{6}\)

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

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

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

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

\(\overrightarrow {{u_1}} .\overrightarrow {{u_2}}  = 0\)

\(\overrightarrow {{u_1}} .\overrightarrow {{u_2}}  = \overrightarrow 0 \)       

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

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

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

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

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

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

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

\(y = {2^x}\)

\(y = 3{x^3}\)

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