Cho tam giác $ABO$ vuông tại $O$, có góc \(\widehat {BAO} = {30^0},AB = a\) . Quay tam giác $ABO$ quanh trục $AO$ ta được một hình nón có diện tích xung quanh bằng:

\(\left\{\begin{array}{l}x = \dfrac{1}{3} - 5t\\y =  - \dfrac{1}{3} - 4t\\z = 3t\end{array}\right.\)

\(\left\{ \begin{array}{l}x = \dfrac{1}{3} + 5t\\y =  - \dfrac{1}{3} - 4t\\z = 3t\end{array} \right.\)

\(\left\{ \begin{array}{l}x = \dfrac{1}{3} + 5t\\y =  - \dfrac{1}{3} + 4t\\z = 3t\end{array} \right.\)

\(\left\{ \begin{array}{l}x = \dfrac{1}{3} - 5t\\y = - \dfrac{1}{3} - 4t\\z =  - 3t\end{array} \right.\)

\({S_{xq}} = \dfrac{1}{3}\pi rl\)

\({S_{xq}} = \pi {r^2}l\)

\({S_{xq}} = \pi rl + \pi {r^2}\)

\({S_{xq}} = \pi rl\)

\(\dfrac{{{a^3}\sqrt 3 }}{2}\)

\(\dfrac{{{a^3}\sqrt 3 }}{3}\)          

\({a^3}\sqrt 3 \)           

\(\dfrac{{3{a^3}\sqrt 3 }}{2}\)

$y = {x^3} + 3x + 2$

$y = {x^3} - 3x + 2$

$y = {x^4} + 3{x^2} + 2$

$y = \dfrac{{x - 1}}{{x + 1}}$

\((S):{(x + 1)^2} + {(y + 1)^2} + {z^2} = 65.\)           

\((S):{(x + 1)^2} + {(y - 1)^2} + {z^2} = 9.\)

\((S):{(x - 1)^2} + {(y + 2)^2} + {z^2} = 64.\)

\((S):{(x + 1)^2} + {(y - 1)^2} + {(z + 2)^2} = 65.\) 

$f'\left( x \right) \ge 0,\forall x \in R.$         

$f'\left( x \right) = 0,\forall x \in R$

$f'\left( x \right) < 0,\forall x \in R.$

$f'\left( x \right) \le 0,\forall x \in R.$

$F\left( x \right) = x\sin x - \cos x + C.$

$F\left( x \right) = {\rm{\;}} - x\sin x - \cos x + C.$

$F\left( x \right) = x\sin x + \cos x + C.$

$F\left( x \right) = {\rm{\;}} - x\sin x + \cos x + C.$