tim m để hàm số y=x^3 - 3(m+1)x^2 +3m(m+2)x +1 đồng biến trên (-vô cùng ;-1) và (2;+ vô cùng) ... help me

\[\begin{array}{l} y = {x^3} - 3\left( {m + 1} \right){x^2} + 3m\left( {m + 2} \right)x + 1\\ \Rightarrow y' = 3{x^2} - 6\left( {m + 1} \right)x + 3m\left( {m + 2} \right).\\ \Rightarrow y' = 0\\ \Leftrightarrow 3{x^2} - 6\left( {m + 1} \right)x + 3m\left( {m + 2} \right) = 0\,\\ \Leftrightarrow {x^2} - 2\left( {m + 1} \right)x + {m^2} + 2m = 0\,\,\,\,\,\left( * \right)\\ TH1:\,\,\,hs\,\,\,DB\,\,tren\,\,\,R\\ \Leftrightarrow y' \geq 0\,\,\,\forall x \in R\\ \Leftrightarrow \Delta ' \leq 0\\ \Leftrightarrow {\left( {m + 1} \right)^2} - {m^2} - 2m \leq 0\\ \Leftrightarrow {m^2} + 2m + 1 - {m^2} - 2m \leq 0\\ \Leftrightarrow 1 \leq 0\,\,\,\left( {vo\,\,ly} \right)\\ \Rightarrow hs\,\,\,k\,\,\,DB\,\,\,tren\,\,\,R.\\ TH2:\,\,pt\,\,y' = 0\,\,co\,\,2\,\,nghiem\,\,pb\\ \Leftrightarrow \Delta ' > 0 \Leftrightarrow {\left( {m + 1} \right)^2} - {m^2} - 2m > 0 \Leftrightarrow 1 > 0\,\,\,\forall m.\\ \Rightarrow pt\,\,\,\left( * \right)\,\,\,co\,\,\,2\,\,nghiem\,\,\,pb\,\,\,{x_1},\,\,{x_2}\,\,\,voi\,\,moi\,\,m.\\ \Rightarrow \left[ \begin{array}{l} {x_1} = m + 1 + 1 = m + 2\\ {x_2} = m + 1 - 1 = m \end{array} \right..\\ Ta\,\,co\,\,\,BXD:\\ \,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,m\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,m + 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \\ \Rightarrow hs\,\,\,DB\,\,\,tren\,\,\,\left( { - \infty ;\,\, - 1} \right)\,\,va\,\,\,\left( {2; + \infty } \right)\\ \Leftrightarrow \left\{ \begin{array}{l} - 1 \le m\\ m + 2 \le 2 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m \ge - 1\\ m \le 0 \end{array} \right. \Leftrightarrow - 1 \le m \le 0.\\ Vay\,\,\, - 1 \le m \le 0\,\,thoa\,\,man\,\,bai\,\,toan. \end{array}\]