\(y=x^3-3x^2+2\)
=>\(y'=3x^2-3\cdot2x=3x^2-6x\)
Đặt y'=0
=>\(3x^2-6x=0\)
=>3x(x-2)=0
=>\(\left[{}\begin{matrix}x=0\\x=2\left(loại\right)\end{matrix}\right.\)
\(y\left(-1\right)=\left(-1\right)^3-3\cdot\left(-1\right)^2+2=-1-3+2=-2\)
\(y\left(0\right)=0^3-3\cdot0^2+2=2\)
\(y\left(1\right)=1^3-3\cdot1^2+2=1-3+2=0\)
=>M=2; m=-2
=>M+m=0
=>Chọn B