Đáp án:
`A`
Giải thích các bước giải:
Gọi $M(x_0;x_0^3-3x_0^2+x_0-5) \in (C)$
$y' = 3x_0^2-6x_0+1$
$\to PTTT : y = (3x_0^2-6x_0+1)(x-x_0) + x_0^3-3x_0^2 +x_0-5$
PTTT $\perp y= \dfrac{1}{2} x+2017$
$\to (3x_0^2-6x_0+1). \dfrac{1}{2} = - 1$
$\to x_0=1$
$\to y= - 2(x-1)-6$
$= - 2x-4$