\(\left(x-3\right)\left(x+1\right)\left(2-3x\right)>0.\)

Vậy \(\left(x-3\right)\left(x+1\right)\left(2-3x\right)>0\) khi \(x\in\left(-\infty;-1\right)\cup\left(\dfrac{2}{3};3\right)\cup\left(3;+\infty\right).\)

Ta có: \(-\dfrac{1}{x-2}\ge0\)

nên x-2<0

hay x<2

\(-\dfrac{1}{x-2}\ge0\Leftrightarrow x-2\le0\Leftrightarrow x\le2\)

Mà : $x ≠ 2 $

Do đó, bất phương trình vô nghiệm

\(\dfrac{-1}{x-2}\ge0\)

ĐKXĐ: \(x-2\ne0\Leftrightarrow x\ne2\)

\(\Leftrightarrow-1\left(x-2\right)\ge0\)

\(\Leftrightarrow x-2\le0\)

\(\Leftrightarrow x\le2\) 

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(không-thõa-mãn-ĐKXĐ\right)\\x< 2\left(thỏa-mãn-ĐKXĐ\right)\end{matrix}\right.\)

Vậy \(S=\left\{x|x< 2\right\}\)

\(a,\left(4x-1\right)\left(x^2+12\right)\left(-x+4\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-1>0\\x^2+12>0\left(LD\forall x\right)\\-x+4>0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x>1\\-x>-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{4}\\x< 4\end{matrix}\right.\)

Vậy \(S=\left\{x|\dfrac{1}{4}< x< 4\right\}\)

\(b,\left(2x-1\right)\left(5-2x\right)\left(1-x\right)< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1< 0\\5-2x< 0\\1-x< 0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{1}{2}\\x>\dfrac{5}{2}\\x< 1\end{matrix}\right.\)

Vậy \(S=\left\{x|1>x>\dfrac{5}{2}\right\}\)

Mn giúp mik vs ạ , cảm ơn nhiều

\(x\left(x-1\right)>0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x>1\\x< 0\end{matrix}\right.\)