Đáp án:
\(m = {5 \over 4}\)
Giải thích các bước giải:
\(\eqalign{
& y = f\left( x \right) = 2{x^3} + 3\left( {m - 1} \right){x^2} + 6\left( {m - 2} \right)x - 1 \cr
& y' = 6{x^2} + 6\left( {m - 1} \right)x + 6\left( {m - 2} \right) = 0 \cr
& \Delta ' = 9{\left( {m - 1} \right)^2} - 36\left( {m - 2} \right) \cr
& \,\,\,\,\,\, = 9\left( {{m^2} - 2m + 1} \right) - 36m + 72 \cr
& \,\,\,\,\,\, = 9{m^2} - 54m + 81 = 9{\left( {m - 3} \right)^2} \cr
& Ham\,\,so\,\,co\,\,2\,\,diem\,\,CT \Rightarrow \Delta > 0 \Rightarrow m \ne 3 \cr
& Ta\,\,co: \cr
& y = y'\left( {{1 \over 3}x + {1 \over 6}\left( {m - 1} \right)} \right) + 4\left( {m - 2} \right)x - {m^2} + 3m - 3 \cr
& \Rightarrow PT\,\,duong\,\,thang\,\,di\,\,qua\,\,2\,\,diem\,\,cuc\,\,tri\,\,la \cr
& \left( d \right):\,\,y = 4\left( {m - 2} \right)x - {m^2} + 3m - 3 \cr
& \left( d \right)//\left( \Delta \right):\,\,y = - 3x + 4 \Leftrightarrow \left\{ \matrix{
4\left( {m - 2} \right) = - 3 \hfill \cr
- {m^2} + 3m - 3 \ne 4 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
4m - 8 = - 3 \hfill \cr
{m^2} - 3m + 7 \ne 0\,\,\left( {luon\,\,dung} \right) \hfill \cr} \right. \Leftrightarrow m = {5 \over 4}\,\,\left( {tm} \right) \cr
& Vay\,\,m = {5 \over 4} \cr} \)
