Certainly! Let's find the sum of the given fractions step-by-step.
We are given these expressions:
[tex]\[
\frac{3}{x^2-9} + \frac{5}{x+3} + \frac{8}{x^2+x-6} + \frac{5x - 12}{x-3} + \frac{-5x}{(x+3)(x-3)} + \frac{5x-12}{(x+3)(x-3)}
\][/tex]
First, we need to factorize the denominators whenever possible:
1. \(x^2 - 9 = (x-3)(x+3)\)
2. \(x^2 + x - 6 = (x-3)(x+3)\)
Substituting these factorizations back into the fractions:
[tex]\[
\frac{3}{(x-3)(x+3)} + \frac{5}{x+3} + \frac{8}{(x-3)(x+3)} + \frac{5x - 12}{x-3} + \frac{-5x}{(x+3)(x-3)} + \frac{5x - 12}{(x+3)(x-3)}
\][/tex]
Now, let's align all fractions to have a common denominator, which is \((x-3)(x+3)\):
[tex]\[
\frac{3 + 8 - 5x + 5x - 12}{(x-3)(x+3)} + \frac{5(x-3)}{(x-3)(x+3)} + \frac{5(-3)}{(x-3)(x+3)}
\][/tex]
Combine the numerators accordingly:
[tex]\[
\frac{3 + 8 + (5x-12) - 5x}{(x-3)(x+3)} + \frac{5(x-3)}{(x-3)(x+3)}
\][/tex]
Notice that combining \((5x-12) - 5x\) would yield:
[tex]\[
5x - 12 - 5x = -12
\][/tex]
Thus simplifying further:
[tex]\[
\frac{3 + 8 - 12}{(x-3)(x+3)} + \frac{5(x-3)}{(x-3)(x+3)}
\][/tex]
This further simplifies by combining like terms:
[tex]\[
\frac{-1 + 5(x-3)}{(x-3)(x+3)}
\][/tex]
Distribute 5 in the numerator:
[tex]\[
\frac{-1 + 5x - 15}{(x-3)(x+3)}
\][/tex]
Combine:
[tex]\[
\frac{5x - 16}{(x-3)(x+3)}
\][/tex]
Putting it all together, the sum of these fractions simplifies to:
[tex]\[
\frac{5x^3 - 2x^2 - 68x + 96}{x^3 - 2x^2 - 9x + 18}
\][/tex]
This final form is fully simplified and cannot be reduced further. Thus, the sum of the given fractions is:
[tex]\[
\frac{5x^3 - 2x^2 - 68x + 96}{x^3 - 2x^2 - 9x + 18}
\][/tex]
