If $\dfrac {(2x+3)(3x-4)}{(x-1)(4x+5)}=f(x)-\dfrac {5}{9(x-1)}+\dfrac {31}{18(4x+5)}$, then $f(x)=$

$\begin{array}{l}\frac{(2 x+3)(3 x-4)}{(x-1)(4 x+5)}=f(x)-\frac{5}{9(x-1)}+\frac{31}{18(4 x+5)} \\ \Rightarrow \frac{(2 x+3)(3 x-4)}{(x-1)(4 x+5)}+\frac{5}{9(x-1)}-\frac{31}{18(4 x+5)}=f(x)\end{array}$

$\begin{array}{l}\Rightarrow \frac{18(2 x+3)(3 x-4)+10(4 x+5)-31(x-1)}{18(x-1)(4 x+5)}=f(x) \\ \Rightarrow \frac{18\left\{6 x^{2}-8 x+9 x-12\right\}+40 x+50-31 x+31}{18(x-1)(4 x+5)}=f(x)\end{array}$

$\Rightarrow \frac{108 x^{2}+18 x-216+40 x+50-31 x+31}{18(x-1)(4 x+5)}=f(x)$

$\begin{array}{l}\Rightarrow \frac{108 x^{2}+27 x-135}{18(x-1)(4 x+5)}=f(x) \\ \Rightarrow \frac{27\left(4 x^{2}+x-5\right)}{18(x-1)(4 x+5)}=f(x)\end{array}$

$\begin{array}{l}\Rightarrow \frac{27\left\{4 x^{2}+5 x-4 x-5\right\}}{18(x-1)(4 x+5)}=f(x) \\ \Rightarrow \frac{3}{2} . \frac{x(4 x+5)-1(4 x+5)}{(x-1)(4 x+5)}=f(x)\end{array}$

$\Rightarrow \frac{3}{2} \cdot \frac{(x-1) \cdot(4 x+5)}{(x-1)(4 x+5)}=f(x)$

$\Rightarrow f(x)=\frac{3}{2}$