$\dfrac{9x}{5} + \dfrac{3x}{5} + \dfrac{x}{5} + ... = \dfrac{27}{4} = 3 \cdot \underbrace{\left(\dfrac{3x}{5} + \dfrac{x}{5} + ...\right)}_{\frac{27}{4} - \frac{9x}{5}}$ $\therefore$ $\dfrac{27}{4} = 3 \cdot \left(\dfrac{27}{4} - \dfrac{9x}{5}\right) =\dfrac{81}{4} - \dfrac{27x}{5} = \dfrac{27}{4}$ $\implies \dfrac{3}{4} - \dfrac{x}{5} = \dfrac{1}{4} \implies \dfrac{x}{5} = \dfrac{1}{2} \implies \boxed{x = \dfrac{5}{2}}$

$$\frac{9x}{5} + \frac{3x}{5} + \frac{x}{5} + \frac{x}{3\cdot 5} + \cdots = \frac{13x}{5} + \frac{x}{5} \cdot \underbrace{\left( \frac{1}{3} + \frac{1}{9} + \cdots\right)}_{\large{\frac{1}{2}}}= \frac{27}{4}$$$$\frac{13x}{5} + \frac{0,5x}{5} = \frac{27}{4} \implies \frac{27x}{10} = \frac{27}{4} \implies \boxed{x = \frac{5}{2}}$$