a/ \(\left(x+1\right)\left(x-2\right)< 0\)
TH1:\(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\) (vô lý)
TH2:\(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\)\(\Rightarrow-1< x< 2\)
Vậy.........
b/ \(\left(x-3\right)\left(x-4\right)>0\)
TH1:\(\left\{{}\begin{matrix}x-3>0\\x-4>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>3\\x>4\end{matrix}\right.\)\(\Rightarrow x>4\)
TH2:\(\left\{{}\begin{matrix}x-3< 0\\x-4< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< 3\\x< 4\end{matrix}\right.\)\(\Rightarrow x< 3\)
Vậy...............
c/ \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}-\dfrac{1}{8}\)
\(\Rightarrow\dfrac{-1}{12}< x< -\dfrac{5}{48}\)
Vậy...............
Để ( x + 1 ) ( x - 2 ) < 0
=> x + 1 và x - 2 phải khác dấu mà x + 1 > x + 2
=> x + 1 dương x + 2 âm
Tức là x + 1 > 0 => x > - 1 và x - 2 < 0 => x < 2
a) \(\left(x+1\right)\left(x-2\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}x+1>0;x-2< 0\\x+1< 0;x-2>0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>-1;x< 2\\x< -1;x>2\end{matrix}\right.\)
\(\Rightarrow-1< x< 2\) (tm)
Vậy \(-1< x< 2.\)
b) \(\left(x-3\right)\left(x-4\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}x-3>0;x-4>0\\x-3< 0;x-4< 0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>3;x>4\\x< 3;x< 4\end{matrix}\right.\)
\(\Rightarrow3< x< 4\)
Vậy \(3< x< 4.\)
c) \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
\(\Rightarrow\dfrac{-1}{12}< x< \dfrac{1}{8}\)
Vậy \(\dfrac{-1}{12}< x< \dfrac{1}{8}.\)
Để ( x- 3 ) ( x - 4 ) > 0 => x - 3 và x - 4 cùng dấu
th1 : x - 3 < 0 => x < 3
x - 4 < 0 => x < 4
th2 : x - 3 > 0 => x > 3
x - 4 > 0 => x > 4
