If the fractional part of $\frac{1}{x}$  and  $x^2$ for some  $x\in (\sqrt{2},\sqrt{3})$ are equal, then the value of  $x^4-\frac{3}{x}=$

 $$\begin{gathered} \sqrt{2}<x<\sqrt{3}\Rightarrow 2<x^2<3 \\ \Rightarrow \frac{1}{\sqrt{2}}>\frac{1}{x}>\frac{1}{\sqrt{3}} \\ \therefore \left[\frac{1}{x}\right]=0 \end{gathered}$$
 Given  $\left\{\frac{1}{x}\right\}=\left\{x^2\right\}\Rightarrow \frac{1}{x}-\left[\frac{1}{x}\right]=x^2-\left[x^2\right]$
 $$\begin{gathered} \Rightarrow \frac{1}{x}-0=x^2-2\Rightarrow x^3-2x=1 \\ \Rightarrow x^3-2x-1=0\Rightarrow x^3+x^2-x^2-x-x-1=0 \\ \Rightarrow (x+1)(x^2-x-1)=0 \end{gathered}$$
 When  $x\ne -1$ then  $x^2-x-1=00\Rightarrow x=1+\frac{1}{x}\Rightarrow -\frac{1}{x}=1-x$
New again  $x^2=x+1\Rightarrow x^4=x^2+1+2x$
 $$\begin{gathered} \because x^4-\frac{3}{x}={(x+1)}^2+3(1-x)=x^2+2x+1+3-3x \\ =x^2-x+4=1+4=5 \end{gathered}$$