\(\left(x+1\right)\left(x-2\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1< 0\Rightarrow x< -1\\x-2>0\Rightarrow x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\Rightarrow x>-1\\x-2< 0\Rightarrow x< 2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-1< x< 2\)
\(\left(x-2\right)\left(x+\dfrac{2}{3}\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\Rightarrow x>2\\x+\dfrac{2}{3}>0\Rightarrow x>-\dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\Rightarrow x< 2\\x+\dfrac{2}{3}< 0\Rightarrow x< -\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x< 2\) hoặc \(x>-\dfrac{2}{3}\)
a, (x+1)(x+2)<0
TH1:x+1>0\(\Rightarrow\)x>-1
x-2<0\(\Rightarrow\)x<2
\(\Rightarrow\)-1<x<2
TH2:x+1<0\(\Rightarrow\)x<-1
x-2>0\(\Rightarrow\)x>2
\(\Rightarrow\)Không có giá trị x thỏa mãn
Câu b cũng như vậy nhưng là cùng dấu bạn nhé!
a)\(\left(x+1\right)\left(x-2\right)< 0\)
=>x+1 và x-2 khác dấu nhau
mà x+1>x-2
=>\(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\Rightarrow-1< x< 2\)
=>\(x\in\left\{0;1\right\}\)
Chúc Bạn Học Tốt!!!