Câu hỏi 11

TXĐ: \(D = \mathbb{R}\).

Ta có: \(y' = {x^2} - 8x - 1\)

Lấy \(y\) chia cho \(y'\) ta có: \(y = \left( {\dfrac{1}{3}x - \dfrac{4}{3}} \right).y' - \dfrac{{34}}{3}x + \dfrac{8}{3}\)

Ta có: \(A\left( {{x_1};{y_1}} \right),\,B\left( {{x_2};{y_2}} \right)\) là hai điểm cực trị \(y'\left( {{x_1}} \right) = y'\left( {{x_2}} \right) = 0 \Rightarrow \left\{ \begin{array}{l}{y_1} =  - \dfrac{{34}}{3}{x_1} + \dfrac{8}{3}\\{y_2} =  - \dfrac{{34}}{3}{x_2} + \dfrac{8}{3}\end{array} \right.\)

Khi đó ta có: \(P = \dfrac{{{y_1} - {y_2}}}{{{x_1} - {x_2}}} = \dfrac{{ - \dfrac{{34}}{3}{x_1} + \dfrac{8}{3} + \dfrac{{34}}{3}{x_2} - \dfrac{8}{3}}}{{{x_1} - {x_2}}} = \dfrac{{ - \dfrac{{34}}{3}\left( {{x_1} - {x_2}} \right)}}{{{x_1} - {x_2}}} =  - \dfrac{{34}}{3}\)

Chọn C.