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Câu hỏi: Gọi $A\left( {{x}_{1}};{{y}_{1}} \right), B\left( {{x}_{2}};{{y}_{2}} \right)$ là hai điểm cực trị của hàm số $y=\dfrac{1}{3}{{x}^{3}}-4{{x}^{2}}-x+4$. Tính $P=\dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}.$
A. $-\dfrac{17}{3}$.
B. $\dfrac{17}{3}$.
C. $\dfrac{34}{3}$.
D. $-\dfrac{34}{3}$.
A. $-\dfrac{17}{3}$.
B. $\dfrac{17}{3}$.
C. $\dfrac{34}{3}$.
D. $-\dfrac{34}{3}$.
$y'={{x}^{2}}-8x-1$. $y'=0\Leftrightarrow \left[ \begin{aligned}
& x=4+\sqrt{17}\Rightarrow y=-\dfrac{128+34\sqrt{17}}{3} \\
& x=4-\sqrt{17}\Rightarrow y=\dfrac{-128+34\sqrt{17}}{3} \\
\end{aligned} \right.$.
Khi đó: $A\left( 4+\sqrt{17};-\dfrac{128+34\sqrt{17}}{3} \right); B\left( 4-\sqrt{17};\dfrac{-128+34\sqrt{17}}{3} \right)$. Vậy $P=-\dfrac{34}{3}$.
& x=4+\sqrt{17}\Rightarrow y=-\dfrac{128+34\sqrt{17}}{3} \\
& x=4-\sqrt{17}\Rightarrow y=\dfrac{-128+34\sqrt{17}}{3} \\
\end{aligned} \right.$.
Khi đó: $A\left( 4+\sqrt{17};-\dfrac{128+34\sqrt{17}}{3} \right); B\left( 4-\sqrt{17};\dfrac{-128+34\sqrt{17}}{3} \right)$. Vậy $P=-\dfrac{34}{3}$.
Đáp án D.