Cho đồ thị hàm số $y=x^3-3x^2+2\,\,(C).$ Gọi $A,\,B$ là hai điểm cực. Tìm tọa độ $M\in(\Delta):y=3x-2$ để $|\overrightarrow{MA}+\overrightarrow{MB}|\,min.$
$y'=3x^2-6x=3x(x-2)$
$y'=0\Rightarrow \left[ {\begin{matrix}x=0\Rightarrow y=2\\ x=2\Rightarrow y=-2 \end{matrix}} \right.$
Gọi $A(0;2), B(2;-2)$
$M\in \Delta \Rightarrow M(3t-2;t)$
$\Rightarrow \begin{cases}\overrightarrow{MA}=(2-3t;2-t)\\ \overrightarrow{MB}=(4-3t;-2-t) \end{cases}\Rightarrow \overrightarrow{MA}+\overrightarrow{MB}=(6-6t;-2t)$
$\Rightarrow \left| {\overrightarrow{MA}+\overrightarrow{MB}} \right|=\sqrt{(6-6t)^2+4t^2}$
mà $(6-6t)^2+4t^2=(2\sqrt{10}t-\frac{9\sqrt{10}}{5})^2+\frac{18}{5}\geq \frac{18}{5}$
$\Rightarrow \Rightarrow \left| {\overrightarrow{MA}+\overrightarrow{MB}} \right|_{min}\Leftrightarrow (2\sqrt{10}t-\frac{9\sqrt{10}}{5})^2=0\Leftrightarrow t=\frac{9}{10}$
$\Rightarrow M(\frac{7}{10};\frac{9}{10})$

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