Cho (C) : y = mx3 - 3mx2 + (2m + 1)x + 3 - m Tìm m để (C) có CĐ, CT. CMR: khi đó đường thẳng đi qua CĐ,CT luôn đi qua 1 điểm cố định.

$$\eqalign{ & \left( C \right):\,\,y = m{x^3} - 3m{x^2} + \left( {2m + 1} \right)x + 3 - m \cr & y' = 3m{x^2} - 6mx + 2m + 1 = 0 \cr & Ham\,\,so\,\,co\,\,2\,\,CT \Leftrightarrow PT\,\,y' = 0\,\,{\mathop{\rm co}\nolimits} \,\,2\,\,nghiem\,\,pb \cr & \Leftrightarrow \left\{ \matrix{ m \ne 0 \hfill \cr \Delta ' = 9{m^2} - 3m\left( {2m + 1} \right) > 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ m \ne 0 \hfill \cr 9{m^2} - 6{m^2} - 3m > 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ m \ne 0 \hfill \cr 3{m^2} - 3m > 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ m > 1 \hfill \cr m < 0 \hfill \cr} \right. \cr & Ta\,\,co:\,\,y = y'\left( {{1 \over 3}x - {1 \over 3}} \right) - {2 \over 3}\left( {m - 1} \right)x - 3m + 2 \cr & \Rightarrow Pt\,\,duong\,\,thang\,\,di\,\,qua\,\,2\,\,CT\,\,la:\,\,y = - {2 \over 3}\left( {m - 1} \right)x - 3m + 2 \cr & \Leftrightarrow - {2 \over 3}mx + {2 \over 3}x - 3m + 2 - y = 0 \cr & \Leftrightarrow \left( { - {2 \over 3}x - 3} \right)m + {2 \over 3}x + 2 - y = 0 \cr & PT\,\,dung\,\,\forall m \Leftrightarrow \left\{ \matrix{ - {2 \over 3}x - 3 = 0 \hfill \cr {2 \over 3}x + 2 - y = 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ x = - {9 \over 2} \hfill \cr y = - 1 \hfill \cr} \right. \Rightarrow Diem\,\,co\,\,dinh:\,\,\left( { - {9 \over 2}; - 1} \right) \cr} $$