Đáp án:
Đáp án C.
Giải thích các bước giải:
\[\begin{array}{l}
y = 3{x^4} - 4{x^3} - 6{x^2} + 12x + 1\\
\Rightarrow y' = 12{x^3} - 12{x^2} - 12x + 12\\
\Rightarrow y' = 0\\
\Leftrightarrow 12{x^3} - 12{x^2} - 12x + 12 = 0\\
\Leftrightarrow {x^3} - {x^2} - x + 1 = 0\\
\Leftrightarrow {x^2}\left( {x - 1} \right) - \left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {{x^2} - 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right){\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 1\,\,\left( {boi\,\,1} \right)\\
x = 1\,\,\,\left( {boi\,\,2} \right)
\end{array} \right.\\
\Rightarrow x = - 1\,\,\,la\,\,diem\,\,\,cuc\,\,\,tri\,\,cua\,\,hs.\\
\Rightarrow M\left( { - 1;\,\, - 10} \right)\\
\Rightarrow x + y = - 11.
\end{array}\]
