Đáp án: $x>-\dfrac{1}{2}$ hoặc $x<-1$ thỏa mãn đề.
Giải thích các bước giải:
Theo phần b) có kết quả : $M = \dfrac{x}{(x+1).(2x+1)}$
Để $M < 1$ thì : $\dfrac{x}{(x+1).(2x+1)} < 1$
$⇔ \dfrac{x}{(x+1).(2x+1)} - 1 < 0 $
$⇔ \dfrac{x-(x+1).(2x+1)}{(x+1).(2x+1)} < 0 $
$⇔ \dfrac{x-2x^2-3x-1}{(x+1).(2x+1)} < 0 $
$⇔ \dfrac{-2x^2-2x-1}{(x+1).(2x+1)} < 0 $ (1)
Ta thấy : $-2x^2-2x-1$
$ = -2.\bigg(x^2+x+\dfrac{1}{2}\bigg)$
$ = -2.\bigg[\bigg(x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}\bigg) + \dfrac{1}{4}\bigg]$
$ = -2.\bigg(x+\dfrac{1}{2}\bigg)^2- \dfrac{1}{2} ≤ - \dfrac{1}{2} < 0 $
Theo $(1)$ thì : $(x+1).(2x+1) > 0 $
$⇔ \left[ \begin{array}{l}\left\{ \begin{array}{l}x+1>0\\2x+1>0\end{array} \right.\\\left\{ \begin{array}{l}x+1<0\\2x+1<0\end{array} \right.\end{array} \right.$
$⇔ \left[ \begin{array}{l}\left\{ \begin{array}{l}x+1>0\\2x+1>0\end{array} \right.\\\left\{ \begin{array}{l}x+1<0\\2x+1<0\end{array} \right.\end{array} \right.$
$⇔ \left[ \begin{array}{l}\left\{ \begin{array}{l}x>-1\\x>-\dfrac{1}{2}\end{array} \right.\\\left\{ \begin{array}{l}x<-1\\x<-\dfrac{1}{2}\end{array} \right.\end{array} \right. $
$⇔\left[ \begin{array}{l}x>-\dfrac{1}{2}\\x<-1\end{array} \right.$
Vậy $x>-\dfrac{1}{2}$ hoặc $x<-1$ thỏa mãn đề.
