=> 2x +1 > x - 3

=> 2x - x > -3 - 1

=> x> -4

\(\frac{2x+1}{x-3}>1\)

\(\Leftrightarrow2x+1>x-3\)

\(\Leftrightarrow2x-x>-3-1\)

\(\Leftrightarrow x>-4\)

\(\frac{2x+1}{x-3}\)>1

<=> 2x+1 > x-3

<=> 2x-x >-3-1

<=> x > -4

Vậy BPT có tập No S= \(\hept{\begin{cases}x\\x>4\end{cases}}\)

( 2x + 1)( 3 - 2x)( 1 - x) > 0

Lập bảng xét dấu , ta có :

Vậy , nghiệm của BPT : \(\dfrac{-1}{2}< x< 1\) hoặc : x > \(\dfrac{3}{2}\)

2( x - 1 ) - 5 = 3( 5 - 3x)

2x - 2 - 5 = 15 - 9x

2x - 7 = 15 - 9x

2x + 9x = 15 + 7

11x = 22

x = 2

Vậy x = 2 

\(2\left(x-1\right)-5=3\left(5-3x\right)\)

\(\Leftrightarrow2x-2-5=15-9x\)

\(\Leftrightarrow2x-\left(2+5\right)=15-9x\)

\(\Leftrightarrow2x-7=15-9x\)

\(\Leftrightarrow2x+9x=15+7\)

\(\Leftrightarrow11x=22\)

\(\Leftrightarrow x=22\div11\)

\(\Leftrightarrow x=2\)

\(\text{Vậy }x=2\)

2( x - 1 ) - 5 = 3( 5 - 3x)

    2x - 2 - 5 = 15 - 9x

         2x - 7 = 15 - 9x

      2x + 9x = 15 + 7

            11x = 22

                x = 2

=>x^4+4x^2+9-4x^3-6x^2+12x<x^4-4x^3-2x^2+15x-3

=>-2x^2+12x+9<-2x^2+15x-3

=>-3x<-12

=>x>4

\(\left(5x-\frac{2}{3}\right)-\frac{2x^2-x}{2}\ge\frac{x\left(1-3x\right)}{3}-\frac{5x}{4}\)

<=> \(\frac{60x-8-6\left(2x^2-x\right)}{12}\ge\frac{4x\left(1-3x\right)-15x}{12}\)

<=> \(60x-8-12x^2+6x\ge4x-12x^2-15x\)

<=> \(47x\ge8\)

<=> \(x\ge\frac{8}{47}\)

ĐK: \(x\ne2\)

\(2x+\dfrac{3}{2x-4}< 3+\dfrac{3}{2x-4}\\ \Leftrightarrow\dfrac{2x.\left(2x-4\right)+3}{2x-4}< \dfrac{3\left(2x-4\right)+3}{2x-4}\\ \Leftrightarrow4x^2-8x+3-6x+12-3< 0\\ \Leftrightarrow4x^2-14x+12< 0\\ \Leftrightarrow\dfrac{3}{2}< x< 2\)

\(bpt\Leftrightarrow\left[\left(x+1\right)^2+3\right]\left(x-1\right)< 0\)

\(\left(x+1\right)^2+3>0\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)