Đáp án:
\(m=2\)
Giải thích các bước giải:
\(\eqalign{
& y = {x^3} - 3m{x^2} + \left( {m - 1} \right)x + 2m \cr
& y' = 3{x^2} - 6mx + m - 1 = 0 \cr
& De\,\,ham\,\,so\,\,co\,\,2\,\,cuc\,\,tri \cr
& \Rightarrow y' = 0\,\,co\,\,2\,\,nghiem\,\,pb \cr
& \Leftrightarrow \Delta ' = {\left( {3m} \right)^2} - 3\left( {m - 1} \right) = 9{m^2} - 3m + 3 > 0\,\,\left( {luon\,\,dung} \right) \cr
& Gia\,\,su\,\,{x_1};\,\,{x_2}\,\,la\,2\,\,nghiem\,pb\,\,cua\,\,pt\,y' = 0 \cr
& \Rightarrow Ham\,so\,co\,\,2\,\,CT\,\,M\left( {{x_1};{y_1}} \right);\,\,N\left( {{x_2};{y_2}} \right) \cr
& G\,\,la\,\,trong\,\,tam\,\,\Delta MNP \cr
& \Rightarrow {x_G} = {{{x_1} + {x_2} + 1} \over 3} = 1 \cr
& \Leftrightarrow {x_1} + {x_2} = 2 \cr
& Ap\,\,dung\,\,DL\,\,Vi - et:\,\,{x_1} + {x_2} = m \cr
& \Rightarrow m = 2 \cr} \)
