If $f:R-\left\{\frac{5}{2}\right\}\to R-\left\{-1\right\}$  defined by $f\left(x\right)=\frac{2x+3}{5-2x}$  then $f^{-1}\left(x\right)$  is

Let $f\left(x\right)=y\Rightarrow x=f^{-1}\left(y\right)$

$y=\frac{2x+3}{5-2x}$

$5y-2xy=2x+3$

$5y-3=2x+2xy$

$5y-3=x\left(2+2y\right)$

$\frac{5y-3}{2+2y}=x=f^{-1}\left(y\right)$

$f^{-1}\left(y\right)=\frac{5y-3}{2+2y}$

$f^{-1}\left(x\right)=\frac{5x-3}{2+2x}$