a: \(\Leftrightarrow\dfrac{-2x+3}{4}+\dfrac{x-5}{3}>-2\)
=>3(-2x+3)+4(x-5)>-24
=>-6x+9+4x-20>-24
=>-2x-11>-24
=>-2x>-13
hay x<13/2
b: \(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)>2\left(2x-1\right)\)
=>(2x-1)2>0
=>2x-1<>0
hay x<>1/2
c: \(3x\left(x-2\right)-20< \left(2x+3\right)\left(2x-3\right)\)
\(\Leftrightarrow4x^2-9>3x^2-6x-20\)
\(\Leftrightarrow x^2+6x+11>0\)(luôn đúng)
\(\dfrac{3-2x}{4}+\dfrac{5-x}{-3}>-2\)
\(\Leftrightarrow\dfrac{3\left(3-2x\right)-4\left(5-x\right)}{12}>\dfrac{-24}{12}\)
\(\Leftrightarrow3\left(3-2x\right)-4\left(5-x\right)>-24\)
\(\Leftrightarrow9-6x-20+4x>-24\)
\(\Leftrightarrow-2x>-13\)
\(\Leftrightarrow x< \dfrac{13}{2}\)
Vậy \(S=\left\{x|x< \dfrac{13}{2}\right\}\)
b.\(4x-2< \left(2x-1\right)\left(2x+1\right)\)
\(\Leftrightarrow4x-2< 4x^2-1\)
\(\Leftrightarrow4x^2-4x+1>0\)
\(\Leftrightarrow\left(2x-1\right)^2>0\)
\(\Leftrightarrow x>\dfrac{1}{2}\)
Vậy \(S=\left\{x|x>\dfrac{1}{2}\right\}\)
c.\(3x\left(x-2\right)-20< \left(2x+3\right)\left(2x-3\right)\)
\(\Leftrightarrow3x^2-6x-20< 4x^2-9\)
\(\Leftrightarrow x^2+6x+11>0\)
Ta có:\(x^2+6x+11=x^2+6x+9+2=\left(x+3\right)^2+2>2>0\) ( đúng )
Vậy \(x\in R\)
