We are asked to add and simplify the expression $\frac{6}{x^2-9x+20} + \frac{7}{x^2-4x}$.

1. Problem Description

We are asked to add and simplify the expression $\frac{6}{x^2-9x+20} + \frac{7}{x^2-4x}$.

2. Solution Steps

First, we factor the denominators:

$x^2 - 9x + 20 = (x-4)(x-5)$

$x^2 - 4x = x(x-4)$

So the expression becomes $\frac{6}{(x-4)(x-5)} + \frac{7}{x(x-4)}$.

To add these fractions, we need a common denominator. The least common denominator is $x(x-4)(x-5)$.

We multiply the first fraction by $\frac{x}{x}$ and the second fraction by $\frac{x-5}{x-5}$:

$\frac{6}{(x-4)(x-5)} \cdot \frac{x}{x} + \frac{7}{x(x-4)} \cdot \frac{x-5}{x-5} = \frac{6x}{x(x-4)(x-5)} + \frac{7(x-5)}{x(x-4)(x-5)}$

Now we can add the numerators:

$\frac{6x + 7(x-5)}{x(x-4)(x-5)} = \frac{6x + 7x - 35}{x(x-4)(x-5)} = \frac{13x - 35}{x(x-4)(x-5)}$

We can expand the denominator if needed:

$x(x-4)(x-5) = x(x^2 - 5x - 4x + 20) = x(x^2 - 9x + 20) = x^3 - 9x^2 + 20x$

So the expression is $\frac{13x - 35}{x^3 - 9x^2 + 20x}$.

We cannot simplify this fraction further, since $13x-35$ cannot be factored, and it does not share any common factors with the denominator.

3. Final Answer

$\frac{13x-35}{x(x-4)(x-5)}$