Ta có: \(\dfrac{x+4}{x+1}=\dfrac{x+1+3}{x+1}=1+\dfrac{3}{x+1}\)
Muốn \(\dfrac{x+4}{x+1}>0\Rightarrow\dfrac{3}{x+1}>0\)
=> 3> x+1 (Điều kiện x # -1)
=>..........
Để \(\dfrac{x+4}{x+1}< 0\) ta có:
TH1: \(\left[{}\begin{matrix}x+4< 0\\x+1< 0\end{matrix}\right.\)\(\Rightarrow-4< x< -1\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)\(\left(TM\right)\)
TH2: \(\left[{}\begin{matrix}x+4< 0\\x+1>0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x< -5\\x>-1\end{matrix}\right.\) \(\left(vôlý\right)\)
Vậy \(x\in\left\{-3;-2\right\}\)
\(\dfrac{x+4}{x+1}< 0\left(1\right)\Rightarrow x+4\) trái dấu với x +1
lại có x +1 < x+4 \(\Rightarrow\) ( 1) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x+4>0\\x+1< 0\end{matrix}\right.\Leftrightarrow-4< x< -1\)
vậy x = \(\left\{-3;-1\right\}\)
