Để \(\frac{2x-1}{x+2}>0\Rightarrow\hept{\begin{cases}2x-1< 0\\x+2< 0\end{cases}}\)hoặc \(\hept{\begin{cases}2x-1>0\\x+2>0\end{cases}}\)

Nếu \(\hept{\begin{cases}2x-1< 0\\x+2< 0\end{cases}\Rightarrow\hept{\begin{cases}2x< 1\\x< 0-2\end{cases}\Rightarrow}\hept{\begin{cases}x< \frac{1}{2}\\x< -2\end{cases}\Rightarrow}x< -2}\)

Nếu \(\hept{\begin{cases}2x-1>0\\x+2>0\end{cases}\Rightarrow\hept{\begin{cases}2x>1\\x>0-2\end{cases}\Rightarrow}\hept{\begin{cases}x>\frac{1}{2}\\x>-2\end{cases}}\Rightarrow x>\frac{1}{2}}\)

Vậy \(\orbr{\begin{cases}x< -2\\x>\frac{1}{2}\end{cases}}\)

b) Để \(\frac{3-x}{3+x}< 0\Rightarrow\hept{\begin{cases}3-x>0\\3+x< 0\end{cases}}\)hoặc \(\hept{\begin{cases}3-x>0\\3+x< 0\end{cases}}\)

Nếu \(\hept{\begin{cases}3-x>0\\3+x< 0\end{cases}\Rightarrow\hept{\begin{cases}x>3\\x< -3\end{cases}\Rightarrow}-3< x< 3}\)

Nếu \(\hept{\begin{cases}3-x< 0\\3+x>0\end{cases}\Rightarrow\hept{\begin{cases}x< 3\\x>-3\end{cases}}}\Rightarrow3< x< -3\Rightarrow x\in\varnothing\)

Vậy \(-3< x< 3\)

a, (x-5).(x-1) >0
<=> x-5>0 và x-1>0
<=> x-5>0
<=> x>5
x-1>0
<=> x>1
Vậy x>5
b, (2x-3).(x+1) <0
<=> 2x-3<0 và x+1<0
2x-3<0 <=> 2x<3 <=> x<2/3
x+1<0 <=> x<-1
Vậy x<2/3
c, 2x² - 3x +1>0
<=> 2x² - 2x- x +1>0
<=>(x-1). (2x-1) >0
<=> x-1>0 và 2x-1>0
x-1>0 <=> x>1
2x-1>0 <=> 2x>1 <=> x>1/2
Vậy x>1/2

a)\(1-2x< 1\)

\(\Leftrightarrow2x>0\)

\(\Leftrightarrow x>0\)

b)\(\left(x-2\right)^2\left(x+1\right)\left(x-4\right)< 0\)

\(\Leftrightarrow\hept{\begin{cases}x\ne2\\\left(x+1\right)\left(x-4\right)< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ne2\\x+1< 0\\x-4>0\end{cases}}\)hoặc \(\hept{\begin{cases}x\ne2\\x+1>0\\x-4< 0\end{cases}}\)

mà \(x+1>x-4\forall x\)

nên \(\hept{\begin{cases}x\ne2\\x+1>0\\x-4< 0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\ne2\\x>-1\\x< 4\end{cases}}\)

hay \(\hept{\begin{cases}x\ne2\\-1< x< 4\end{cases}}\)

c)\(x-2< 0\)

\(\Leftrightarrow x< 2\)

d)\(\frac{x^2\left(x-3\right)}{x-9}< 0\left(x\ne9\right)\)

\(\Leftrightarrow\hept{\begin{cases}x\ne0\\\frac{x-3}{x-9}< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x-3< 0\\x-9>0\end{cases}}\)hoặc \(\hept{\begin{cases}x\ne0\\x-3>0\\x-9< 0\end{cases}}\)

mà \(x-3>x-9\forall x\)

\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x-3>0\\x-9< 0\end{cases}}\)\(\Leftrightarrow3< x< 9\)

e)\(\frac{5}{x}< 1\left(x\ne0\right)\)

\(\Leftrightarrow x>5\)

f)\(8x>2x\)

\(\Leftrightarrow6x>0\)

\(\Leftrightarrow x>0\)

g)\(x+a< a\)

\(\Leftrightarrow x< 0\)

h)\(x^3< x^2\)

\(\Leftrightarrow x^2\left(x-1\right)< 0\)

\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x-1< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x< 1\end{cases}}\)

a) \(\frac{-13}{2x+1}< 0\)

\(=>2x+1>0\)

\(=>2x>-1\)

\(=>x=\frac{1}{2}\)

b) \(\frac{x-1}{x+3}>0\)

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

c) \(\frac{2x+2}{x-4}< 0\)

\(=>2x+2< 0=>x< -1\)