Solve the following equation by factorisation:

$\dfrac{1}{2a + b + 2x} = \dfrac{1}{2a} + \dfrac{1}{b} + \dfrac{1}{2x}$

Given,

$\dfrac{1}{2a + b + 2x} = \dfrac{1}{2a} + \dfrac{1}{b} + \dfrac{1}{2x} \\[1em] \Rightarrow \dfrac{1}{2a + b + 2x} - \dfrac{1}{2x} = \dfrac{1}{2a} + \dfrac{1}{b} \\[1em] \Rightarrow \dfrac{2x - (2a + b + 2x)}{(2a + b + 2x)(2x)} = \dfrac{b + 2a}{2ab} \\[1em] \Rightarrow \dfrac{ -(2a + b )}{(2a + b + 2x)(2x)} = \dfrac{b + 2a}{2ab}\\[1em] \Rightarrow \dfrac{ -1}{(2a + b + 2x)(2x)} = \dfrac{1}{2ab}\\[1em] \Rightarrow -2ab = (2a + b + 2x)(2x) \\[1em] \Rightarrow -2ab = 4ax + 2bx + 4x^2 \\[1em] \Rightarrow -ab = 2ax + bx + 2x^2 \text{ (Dividing the complete equation by 2) } \\[1em] \Rightarrow 2ax + bx + 2x^2 + ab = 0 \\[1em] \Rightarrow 2x^2 + 2ax + bx + ab = 0 \\[1em] \Rightarrow 2x(x + a) + b(x + a) = 0 \\[1em] \Rightarrow (2x + b)(x + a) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow 2x + b = 0 \text{ or } x + a = 0 \text{ (Zero-product rule) } \\[1em] x = -\dfrac{b}{2} \text{ or } x = -a \\[1em]$

Hence, the roots of given equation are $-\dfrac{b}{2}$, -a.