Đáp án:
Giải thích các bước giải:
$$\eqalign{
& y = \left( {m + 1} \right){x^3} + {x^2} + \left( {2{m^2} + 1} \right)x - 3m + 2\,\,db/R \cr
& y' = 3\left( {m + 1} \right){x^2} + 2x + 2{m^2} + 1 \cr
& TH1:\,\,m + 1 = 0 \Leftrightarrow m = - 1 \cr
& \Rightarrow y' = 2x + 3 > 0 \Leftrightarrow x > - {3 \over 2} \cr
& \Rightarrow Hs\,\,db/\left( { - {3 \over 2}; + \infty } \right) \Rightarrow Loai \cr
& TH2:\,\,m + 1 \ne 0 \Leftrightarrow m \ne - 1 \cr
& Hs\,\,db/R \Leftrightarrow \left\{ \matrix{
m + 1 > 0 \hfill \cr
\Delta ' = 1 - 3\left( {m + 1} \right)\left( {2{m^2} + 1} \right) < 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
m > - 1 \hfill \cr
1 - 3\left( {2{m^3} + m + 2{m^2} + 1} \right) < 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
m > - 1 \hfill \cr
- 6{m^3} - 3{m^2} - 3m - 2 < 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
m > - 1 \hfill \cr
m > - 0,6 \hfill \cr} \right. \Leftrightarrow m > - 0,6 \cr} $$
Đề bài lẻ quá.
