Đáp án:

\(m \in \left( { - \infty ; - 1} \right) \cup \left( {1; + \infty } \right)\)

Giải thích các bước giải:

\(\eqalign{ & y = {1 \over 3}{x^3} - m{x^2} - x + m + {2 \over 3} \cr & Xet\,\,PTHDGD:\,\,{1 \over 3}{x^3} - m{x^2} - x + m + {2 \over 3} = 0 \cr & \Leftrightarrow {x^3} - 3m{x^2} - 3x + 3m + 2 = 0\,\,\left( 1 \right) \cr & \Leftrightarrow \left( {x - 1} \right)\left( {{x^2} + \left( {1 - 3m} \right)x - 2 - 3m} \right) = 0 \cr & \Leftrightarrow \left[ \matrix{ {x_1} = 1 \hfill \cr {x^2} + \left( {1 - 3m} \right)x - 2 - 3m = 0\,\,\left( 2 \right) \hfill \cr} \right. \cr & De\,DTHS\,\,cat\,\,truc\,\,hoanh\,\,tai\,\,3\,\,diem\,\,pb \cr & \Rightarrow \left( 1 \right)\,\,co\,\,3\,\,nghiem\,\,pb \cr & \Rightarrow \left( 2 \right)\,\,co\,\,2\,\,nghiem\,\,pb\,\,khac\,\,\,1 \cr & \Rightarrow \left\{ \matrix{ \Delta = {\left( {1 - 3m} \right)^2} + 4\left( {2 + 3m} \right) > 0 \hfill \cr 1 + 1 - 3m - 2 - 3m \ne 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ 9{m^2} - 6m + 1 + 8 + 6m > 0 \hfill \cr - 6m \ne 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ 9{m^2} + 9 > 0\,\,\left( {luon\,dung} \right) \hfill \cr m \ne 0 \hfill \cr} \right. \Leftrightarrow m \ne 0 \cr & Goi\,\,{x_2},\,\,{x_3}\,\,la\,\,2\,\,nghiem\,\,pb\,\,cua\,\,\left( 2 \right) \cr & Ap\,\,dung\,\,DL\,\,Vi - et\,\,ta\,\,co:\,\,\left\{ \matrix{ {x_1} + {x_2} = 3m - 1 \hfill \cr {x_1}{x_2} = - 2 - 3m \hfill \cr} \right. \cr & Theo\,\,bai\,\,ra\,\,ta\,\,co: \cr & x_1^2 + x_2^2 + x_3^2 > 15 \cr & \Leftrightarrow 1 + {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} > 15 \cr & \Leftrightarrow {\left( {3m - 1} \right)^2} - 2\left( { - 2 - 3m} \right) > 14 \cr & \Leftrightarrow 9{m^2} - 6m + 1 + 4 + 6m > 14 \cr & \Leftrightarrow 9{m^2} > 9 \cr & \Leftrightarrow \left[ \matrix{ m > 1 \hfill \cr m < - 1 \hfill \cr} \right.\,\,\left( {tm\,\,m \ne 0} \right) \cr & Vay\,\,m \in \left( { - \infty ; - 1} \right) \cup \left( {1; + \infty } \right) \cr} \)