a) \(\dfrac{2x-5}{3x+6}>0\)
TH1:
\(\left\{{}\begin{matrix}2x-5>0\\3x+6>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{5}{2}\\x>-2\end{matrix}\right.\)
\(\Leftrightarrow x>\dfrac{5}{2}\)
TH2:
\(\left\{{}\begin{matrix}2x-5< 0\\3x+6< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x< -2\end{matrix}\right.\)
\(\Leftrightarrow x< -2\)
Vậy \(x>\dfrac{5}{2}\) hoặc x < -2 thì \(\dfrac{2x-5}{3x+6}>0\)
b) \(\dfrac{4x-1}{x+3}< 0\)
TH1:
\(\left\{{}\begin{matrix}4x-1< 0\\x+3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{4}\\x>-3\end{matrix}\right.\Leftrightarrow-3< x< \dfrac{1}{4}\)
Vì x là số nguyên \(\Rightarrow x=\left\{-2;-1;0\right\}\)
TH2:
\(\left\{{}\begin{matrix}4x-1>0\\x+3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{4}\\x< -3\end{matrix}\right.\left(L\right)\)
Vậy x = {-2;-1;0} thì\(\dfrac{4x-1}{x+3}< 0\)
