Using properties of proportion, solve for x. Given that x is positive.

$\begin{matrix} \text{(i)} & \dfrac{3x + \sqrt{9x^2 - 5}}{3x - \sqrt{9x^2 - 5}} = 5 \\[0.5em] \text{(ii)} & \dfrac{2x + \sqrt{4x^2 - 1}}{2x - \sqrt{4x^2 - 1}} = 4 \end{matrix}$

(i) Given,

$\dfrac{3x + \sqrt{9x^2 - 5}}{3x - \sqrt{9x^2 - 5}} = \dfrac{5}{1}.$

By componendo and dividendo,

$\Rightarrow \dfrac{3x + \sqrt{9x^2 - 5} + 3x - \sqrt{9x^2 - 5}}{3x + \sqrt{9x^2 - 5} - 3x + \sqrt{9x^2 - 5}} = \dfrac{5 + 1}{5 - 1} \\[1em] \Rightarrow \dfrac{6x}{2\sqrt{9x^2 - 5}} = \dfrac{6}{4} \\[1em] \Rightarrow \dfrac{x}{\sqrt{9x^2 - 5}} = \dfrac{1}{2}$

Squaring both sides we get,

$\dfrac{x^2}{9x^2 - 5} = \dfrac{1}{4} \\[0.5em] \Rightarrow 4x^2 = 9x^2 - 5 \\[0.5em] \Rightarrow 5x^2 - 5 = 0 \\[0.5em] \Rightarrow 5(x^2 - 1) = 0 \\[0.5em] \Rightarrow (x + 1)(x - 1) = 0 \\[0.5em] \Rightarrow x = 1, -1.$

Since, x is positive, hence, x ≠ -1.

Hence, the required value of x is 1.

(ii) Given,

$\dfrac{2x + \sqrt{4x^2 - 1}}{2x - \sqrt{4x^2 - 1}} = \dfrac{4}{1}.$

By componendo and dividendo,

$\Rightarrow \dfrac{2x + \sqrt{4x^2 - 1} + 2x - \sqrt{4x^2 - 1}}{2x + \sqrt{4x^2 - 1} - 2x + \sqrt{4x^2 - 1}} = \dfrac{4 + 1}{4 - 1} \\[1em] \Rightarrow \dfrac{4x}{2\sqrt{4x^2 - 1}} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{2x}{\sqrt{4x^2 - 1}} = \dfrac{5}{3}$

Squaring both sides we get,

$\dfrac{4x^2}{4x^2 - 1} = \dfrac{25}{9} \\[1em] \Rightarrow 4x^2 \times 9 = 25(4x^2 - 1) \\[1em] \Rightarrow 36x^2 = 100x^2 - 25 \\[1em] \Rightarrow 64x^2 = 25 \\[1em] \Rightarrow x^2 = \dfrac{25}{64} \\[1em] \Rightarrow x = \sqrt{\dfrac{25}{64}} \\[1em] \Rightarrow x = \dfrac{5}{8} \text{ or } -\dfrac{5}{8} \\[1em]$

Since, x is positive, hence, x ≠ $-\dfrac{5}{8}.$

Hence, the required value of x is $\dfrac{5}{8}$.