Đáp án:
$\begin{array}{l}
{y_0} = 3\\
\Rightarrow 2x_0^3 - 3{x_0} + 3 = 3\\
\Rightarrow 2x_0^3 - 3{x_0} = 0\\
\Rightarrow \left[ \begin{array}{l}
{x_0} = 0 \Rightarrow y' = 6x_0^2 - 3 = - 3\\
{x_0} = \dfrac{{\sqrt 6 }}{2} \Rightarrow y' = 6x_0^2 - 3 = 6\\
{x_0} = - \dfrac{{\sqrt 6 }}{2} \Rightarrow y' = 6x_0^2 - 3 = 6
\end{array} \right.\\
\Rightarrow PTTT:y = {y_0}'\left( {x - {x_0}} \right) + {y_0}\\
\Rightarrow \left[ \begin{array}{l}
y = - 3.x + 3\\
y = 6.\left( {x - \dfrac{{\sqrt 6 }}{2}} \right) + 3 = 6x - 3\sqrt 6 + 3\\
y = 6.\left( {x + \dfrac{{\sqrt 6 }}{2}} \right) + 3 = 6x + 3\sqrt 6 + 3
\end{array} \right.
\end{array}$
