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 \(S=\left\{x|x< 2\right\}\)

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

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

\(bpt\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\ge0\\4-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1\le0\\4-x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge1\\x< 4\end{matrix}\right.\\\left\{{}\begin{matrix}x\le1\\x>4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow1\le x< 4\)

Vậy .......

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\ge0\\4-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1\le0\\4-x< 0\end{matrix}\right.\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge1\\x< 4\end{matrix}\right.\\\left\{{}\begin{matrix}x\le1\\x>4\end{matrix}\right.\end{matrix}\right.\)

Vậy....

1-3x<0

<=> -3x<-1

<=> x> -1/-3

<=> x> 1/3

1-3x<0

<=> -3x<-1

<=> x<\(\frac{-1}{-3}\)

<=> x<\(\frac{1}{3}\)

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

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

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

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

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

Ta có: \(2x^2-5x+5=2\left(x^2-2.\dfrac{5}{4}x+\dfrac{25}{16}\right)+\dfrac{15}{8}=2\left(x-\dfrac{5}{4}\right)^2+\dfrac{15}{8}>0\)

Lời giải:

Ta có: \(\frac{1}{x(x+1)}< 0\Leftrightarrow x(x+1)< 0\)

\(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x>0\\ x+1< 0\end{matrix}\right.\\ \left\{\begin{matrix} x< 0\\ x+1>0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} 0< x< -1(\text{vô lý})\\ 0> x> -1\end{matrix}\right.\)

\(\Rightarrow 0> x> -1\)