If terms independent of x in the expansion of ${(3x-\frac{1}{x})}^{20}and{(x+\frac{\sqrt[9]{3^{10}}}{x})}^{18}$ are A and B respectively, then $(\frac{9}{38}A+B)$ equals:

Term independent of x in expansion of ${(3x-\frac{1}{x})}^{20}$

$T_{r+1}={}^{20}{C}_r{\left(3x\right)}^{20-r}{\left(\frac{-1}{x}\right)}^r$

When r =10,  $A=T_{11}={}^{20}{C}_{10}{3}^{10} \dots ..\left(1\right)$

Term independent of x in expansion of ${(x+\frac{\sqrt[9]{3^{10}}}{x})}^{18}$

$T_{r+1}={}^{18}{C}_r{\left(x\right)}^{18-r}{\left(\frac{{\left(3^{10}\right)}^{1/9}}{x}\right)}^r$

When r = 9, $B=T_{10}={}^{18}{C}_9{3}^{10}$

So, $(\frac{9}{37}A+B)=(\frac{9}{38}\times {}^{20}{C}_{10}\times 3^{10}+{}^{18}{C}_9\times 3^{10})$   [From eqns. (1) and (2)]

$=(\frac{9}{38}\times \frac{20}{10}\times \frac{19}{9}\times {}^{18}{C}_8\times 3^{10}{+}^{18}{C}_9\times 3^{10})=3^{10}\times {}^{19}{C}_9$