Đáp án:
m<3/2
Giải thích các bước giải:
$\begin{array}{l}
y = \frac{{{X^3}}}{3} - (m - 2){x^2} + (4m - 8)x + m + 1\,co\,2\,cuc\,tri\,{x_1},{x_2}\\
\to y' = {x^2} - 2(m - 2)x + 4m - 8 = 0\,\,co\,2\,nghiem\,phan\,biet\\
\to ' = {(m - 2)^2} - 4m + 8 = {m^2} - 8m + 12 > 0\\
\to m > 6\,\,hoac\,\,m < 2(1)\\
ma\,{x_1} < - 2 < {x_2}\\
nen\,({x_1} + 2)({x_2} + 2) < 0\\
\to {x_1}.{x_2} + 2({x_1} + {x_2}) + 4 < 0\\
\to 4m - 8 + 2.2.(m - 2) + 4 < 0\\
\to 8m - 12 < 0\\
\to m < \frac{3}{2}(2)\\
Tu\,(1)\,va\,(2) \to m < \frac{3}{2}\\
\end{array}$
