a,
\(\left(x+1\right)\left(x-3\right)< 0\)
\(\Rightarrow x+1\text{ và }x-3\text{ khác dấu và }x+1\ne0,x-3\ne0\Rightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x-3< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>1\\x< 3\end{matrix}\right.\Rightarrow1< x< 3\\\left\{{}\begin{matrix}x+1< 0\\x-3>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< -1\\x>3\end{matrix}\right.\Rightarrow\text{mâu thuẫn}\end{matrix}\right.\)
Vậy \(1< x< 3\) thì \(\left(x+1\right)\left(x-3\right)< 0\)
b,
\(\dfrac{x+1}{x-4}>0\)
\(\Rightarrow x+1\text{ và }x-4\text{ cùng dấu và }x+1\ne0,x-4\ne0\Rightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne4\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x-4>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>-1\\x>4\end{matrix}\right.\Rightarrow x>4\\\left\{{}\begin{matrix}x+1< 0\\x-4< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< -1\\x< 4\end{matrix}\right.\Rightarrow x< -1\end{matrix}\right.\)
Vậy khi \(x>4\) hoặc \(x< -1\) thì \(\dfrac{x+1}{x-4}>0\)
\(\left(x+1\right)\left(x-3\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\Rightarrow x>-1\\x-3< 0\Rightarrow x< 3\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\Rightarrow x< -1\\x-3>0\Rightarrow x>3\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-1< x< 3\)
\(\dfrac{x+1}{x-4}>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\Rightarrow x>-1\\x-4>0\Rightarrow x>4\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\Rightarrow x< -1\\x-4< 0\Rightarrow x< 4\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x>-1;x< 4\)
a) Để (x+1)(x-3) <0
Thì ta có 2 TH:
+x+1>0 ; x-3<0
+x+1<0 ; x-3>0
Th1:x+1>0 ; x-3<0
+x+1>0=>x>-1
+x-3<0=>x<3
=>3>x>-1(TM)
Th2:x+1<0 ; x-3>0
+x+1<0=>x<-1
+x-3>0=>x>3
=>-1>x>3 (Loại vì vô lí)
Vậy (x+1).(x-3)<0 <=>3>x>-1
b)Để x+1/x-4 >0
=>x+1>x-4 và x+1; x-4 đều cùng lớn hơn hoặc bé hơn 0
Tự làm tiếp nhá
