If $x\in \mathbf{N}$ and $\frac{(x+2)!}{(2x-1)!}\cdot \frac{(2x+1)!}{(x+3)!}=\frac{72}{7},$then $x$is equal to 

 

$\frac{(2x+1)\left(2x\right)}{x+3}=\frac{72}{7}$

$\begin{array}{l}\Rightarrow 7\left(2x^2+x\right)=36(x+3)\\ \Rightarrow 14x^2-29x-108=0\\ \Rightarrow (14x+27)(x-4)=0\Rightarrow x=4\end{array}$