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

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

=>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

=> 2x +1 > x - 3

=> 2x - x > -3 - 1

=> x> -4

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

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

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

\(\Leftrightarrow x>-4\)

\(\dfrac{1}{3-5x}>\dfrac{1}{2x+3}\)

⇔ \(\dfrac{1}{3-5x}-\dfrac{1}{2x+3}>0\)

⇔ \(\dfrac{2x+3+5x-3}{\left(3-5x\right)\left(2x+3\right)}>0\)

⇔\(\dfrac{7x}{\left(3-5x\right)\left(2x+3\right)}>0\)

Lập bảng xét dấu , ta có :

Vậy , nghiệm của BPT là : x < \(\dfrac{-3}{2}\) hoặc : 0 < x < \(\dfrac{3}{5}\)

\(\dfrac{1}{3-5x}>\dfrac{1}{2x+3}\)

DKXD : \(x\ne\dfrac{3}{5};x\ne\dfrac{-3}{2}\)

\(\Leftrightarrow\dfrac{1}{3-5x}-\dfrac{1}{2x+3}>0\)

\(\Leftrightarrow\dfrac{2x+3}{\left(3-5x\right)\left(2x+3\right)}-\dfrac{3-5x}{\left(3-5x\right)\left(2x+3\right)}>0\)

\(\Leftrightarrow\dfrac{2x+3-3+5x}{\left(3-5x\right)\left(2x+3\right)}>0\)

\(\Leftrightarrow\dfrac{7x}{\left(3-5x\right)\left(2x+3\right)}>0\)

\(\Leftrightarrow7x>0\)

\(\Leftrightarrow x>0\)

Vậy bpt có nghiệm khi \(x>0\) tm \(x\ne\dfrac{3}{5};x\ne\dfrac{-3}{2}\)

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

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

<=> \(60x-8-12x^2+6x\ge4x-12x^2-15x\)

<=> \(47x\ge8\)

<=> \(x\ge\frac{8}{47}\)