\(x\left(x-\dfrac{1}{7}\right)\left(x+\dfrac{1}{9}\right)< 0\)
Vậy phải có 1 số lẻ các số âm
Vậy \(\left\{{}\begin{matrix}x< 0\\x-\dfrac{1}{7}< 0\Rightarrow x< \dfrac{1}{7}\\x+\dfrac{1}{9}< 0\Rightarrow x< -\dfrac{1}{9}\end{matrix}\right.\)
Vậy \(x< \dfrac{1}{7}\)
Vì \(x-\dfrac{1}{7}< x< \dfrac{1}{9}+x\) nên
\(\left\{{}\begin{matrix}x-\dfrac{1}{7}< 0\Rightarrow x< \dfrac{1}{7}\\x>0\\\dfrac{1}{9}+x>0\Rightarrow x>-\dfrac{1}{9}\end{matrix}\right.\)
Vậy \(-\dfrac{1}{9}< x< \dfrac{1}{7}\)
\(\dfrac{4-x}{2x-\dfrac{1}{5}}>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4-x>0\Rightarrow x< 4\\2x-\dfrac{1}{5}>0\Rightarrow x>\dfrac{1}{10}\end{matrix}\right.\\\left\{{}\begin{matrix}4-x< 0\Rightarrow x>4\\2x-\dfrac{1}{5}< 0\Rightarrow x< \dfrac{1}{10}\end{matrix}\right.\end{matrix}\right.\)
Vậy\(\dfrac{1}{10}< x< 4\)
