Đáp án:
$\begin{array}{l}
y = {x^3} + \left( {m + 1} \right){x^2} - \left( {2{m^2} + 3m + 2} \right)x + 1\\
y' = 3{x^2} + 2\left( {m + 1} \right)x - 2{m^2} - 3m - 2\\
có\,\,\Delta ' = {\left( {m + 1} \right)^2} + 6{m^2} + 9m + 6 = 7{m^2} + 11m + 7 > 0\forall m\\
\Rightarrow pt\,y' = 0\,luôn\,có\,2\,nghiệm\,pb\,\,{x_1};{x_2}\\
đề\,bài\,yêu\,cầu\,y' > 0\,\,\forall x > 2\\
\Rightarrow {x_1} < {x_2} < 2\\
\Rightarrow \left( {{x_1} - 2} \right)\left( {{x_2} - 2} \right) > 0\\
\Rightarrow {x_1}{x_2} - 2\left( {{x_1} + {x_2}} \right) + 4 > 0\\
\Rightarrow \frac{{ - 2{m^2} - 3m - 2}}{3} - 2\left( {\frac{{ - 2m - 2}}{3}} \right) + 4 > 0\\
\Rightarrow - 2{m^2} - 3m - 2 + 4m + 4 + 12 > 0\\
\Rightarrow \frac{{1 - \sqrt {113} }}{4} < m < \frac{{1 + \sqrt {113} }}{4}
\end{array}$
