Add $ \dfrac{x-25}{5x-25} $ and $ \dfrac{3x+5}{x^2-5x} $ to get $ \dfrac{ \color{purple}{ x^2-10x+25 } }{ 5x^2-25x }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ x }$ and the second by $\color{blue}{ 5 }$.

$$ \begin{aligned} \frac{x-25}{5x-25} + \frac{3x+5}{x^2-5x} & = \frac{ \left( x-25 \right) \cdot \color{blue}{ x }}{ \left( 5x-25 \right) \cdot \color{blue}{ x }} + \frac{ \left( 3x+5 \right) \cdot \color{blue}{ 5 }}{ \left( x^2-5x \right) \cdot \color{blue}{ 5 }} = \\[1ex] &=\frac{ \color{purple}{ x^2-25x } }{ 5x^2-25x } + \frac{ \color{purple}{ 15x+25 } }{ 5x^2-25x }=\frac{ \color{purple}{ x^2-10x+25 } }{ 5x^2-25x } \end{aligned} $$

Simplify $ \dfrac{x^2-10x+25}{5x^2-25x} $ to $ \dfrac{x-5}{5x} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-5}$.

$$ \begin{aligned} \frac{x^2-10x+25}{5x^2-25x} & =\frac{ \left( x-5 \right) \cdot \color{blue}{ \left( x-5 \right) }}{ 5x \cdot \color{blue}{ \left( x-5 \right) }} = \\[1ex] &= \frac{x-5}{5x} \end{aligned} $$