$\begin{align} & y'=3{{x}^{2}}-6\left( m+1 \right)x+9 \\ & y'=0\Leftrightarrow {{x}^{2}}-2\left( m+1 \right)x+3=0 \\ \end{align}$

Để hs có 2 cực trị

$\Delta '=$${{m}^{2}}+2m-2>0$

$\Leftrightarrow \left[ \begin{align} & m<-1-\sqrt{3} \\ & m>-1+\sqrt{3} \\ \end{align} \right.$

Theo đl Viet, ta được:

$\begin{align} & {{x}_{1}}+{{x}_{2}}=2\left( m+1 \right) \\ & {{x}_{1}}.{{x}_{2}}=3 \\ \end{align}$

$\begin{align} & \left| {{x}_{1}}-{{x}_{2}} \right|=2 \\ & \Leftrightarrow x_{1}^{2}+x_{2}^{2}-2{{x}_{1}}{{x}_{2}}=4 \\ \end{align}$

$\begin{align} & \Leftrightarrow 4{{\left( m+1 \right)}^{2}}-12-4=0 \\ & \Leftrightarrow {{\left( m+1 \right)}^{2}}=4 \\ \end{align}$

$\Leftrightarrow \left[ \begin{align} & m+1=2 \\ & m+1=-2 \\ \end{align} \right.$

$\Leftrightarrow \left[ \begin{align} & m=1\left( nhan \right) \\ & m=-3\left( nhan \right) \\ \end{align} \right.$