Đáp án:
\(\, - {7 \over 2} < m < - 3.\)
Giải thích các bước giải:
$$\eqalign{
& y = {1 \over 3}{x^3} + \left( {m + 3} \right){x^2} + 4\left( {m + 3} \right)x + {m^3} - m \cr
& y' = {x^2} + 2\left( {m + 3} \right)x + 4\left( {m + 3} \right) = 0 \cr
& Ham\,\,so\,\,co\,\,2\,\,CT \Leftrightarrow Pt\,\,y' = 0\,\,co\,\,2\,\,nghiem\,pb. \cr
& \Delta ' = {\left( {m + 3} \right)^2} - 4\left( {m + 3} \right) > 0 \cr
& \Leftrightarrow \left[ \matrix{
m + 3 > 4 \hfill \cr
m + 3 < 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
m > 1 \hfill \cr
m < - 3 \hfill \cr} \right. \cr
& - 1 < {x_1} < {x_2} \Leftrightarrow 0 < {x_1} + 1 < {x_2} + 1 \cr
& \Leftrightarrow \left\{ \matrix{
{x_1} + {x_2} > - 2 \hfill \cr
\left( {{x_1} + 1} \right)\left( {{x_2} + 1} \right) > 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
{x_1} + {x_2} > - 2 \hfill \cr
{x_1}{x_2} + {x_1} + {x_2} + 1 > 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
- 2m - 6 > - 2 \hfill \cr
4m + 12 - 2m - 6 + 1 > 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
2m < - 4 \hfill \cr
2m > - 7 \hfill \cr} \right. \Leftrightarrow - {7 \over 2} < m < - 2 \cr
& Vay\,\, - {7 \over 2} < m < - 3. \cr} $$
