The solution set of the inequality  $|x+2|-|x-1|<x-\frac{3}{2}$ is 

The inequality is $|x+2|-|x-1|<x-\frac{3}{2}$.

Dividing the problem into three intervals :

$\text{ (i) If }x<-2,$ then

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

But $-\frac{3}{2}>-2$ hence no common values 

$\Rightarrow x\in \varphi$

If $-2\le x<1$, then

$$\begin{gathered} (x+2)+(x-1)<x-\frac{3}{2} \\ \Rightarrow x<-\frac{5}{2} \end{gathered}$$

But $-\frac{5}{2}<-2$, hence no common values 

$\Rightarrow x\in \varphi$

$(x+2)-(x-1)<x-\frac{3}{2}\Rightarrow x>\frac{9}{2}$

$\because \frac{9}{2}>1\Rightarrow$common solution is

$x>\frac{9}{2}\Rightarrow x\in \left(\frac{9}{2},\mathrm{\infty }\right)$

$\therefore$ Solution set is$x\in \left(\frac{9}{2},\mathrm{\infty }\right)$