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

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

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