Given

$\rm \frac{b}{x^2 - 5x} - \frac{x}{5x - 25} = - \frac{x + 5}{5x}$

$\rm or, \frac{b}{x(x - 5)} - \frac{x}{5(x - 5)} = - \frac{x + 5}{5x}$

$\rm or, \frac{b}{x (x - 5)} = - \frac{x + 5}{5x} + \frac{x}{5 (x - 5)}$

$\rm or, \frac{b}{x (x - 5)} = - \frac{x + 5}{5x} \cdot \frac{(x - 5)}{(x - 5)} + \frac{x}{5 (x - 5)} \cdot \frac{x}{x}$

$\rm or, \frac{b}{x (x - 5)} = - \frac{ (x - 5)(x + 5) }{5x ( x - 5)} + \frac{x \cdot x}{5 (x - 5) \cdot x}$

$\rm or, \frac{b}{x (x - 5)} = - \frac{ x^2 - 25 }{5x ( x - 5)} + \frac{ x^2 }{5x (x - 5)}$

$\rm or, \frac{b}{x (x - 5)} = \frac{ - (x^2 - 25) + x^2 }{5x ( x - 5) }$

$\rm or, \frac{b}{x (x - 5)} = \frac{ - x^2 + 25 + x^2}{5x (x - 5)}$

$\rm or, \frac{b}{x (x - 5)} = \frac{ 25}{5x (x - 5)}$

$\rm or, \frac{b}{x ( x - 5)} = \frac{ 5 \cdot 5}{5x (x - 5)}$

$\rm or, \frac{b}{x (x - 5)} = \frac{5}{x (x - 5)}$

The expressions in the denominators of both sides of the equation are equal. So, the numerators of these expressions must be equal to hold the equality.

$\rm \therefore b = 5$

Hence, the required value of b is 5.