We have to solve 3/(x+3) + 2/(3-x) + (2x+24)/(x^2-9)

3/(x+3) + 2/(3-x) + (2x+24)/(x^2-9)

=> 3/(x+3) - 2/(x-3) + (2x+24)/(x^2-9)

Make the denominator the same in all the terms

=> [3(x - 3) - 2(3 + x) + (2x + 24)]/(x^2 - 9)

=> [3x - 9 - 6 - 2x + 2x + 24]/(x^2 - 9)

=> [3x + 9]/(x^2 - 9)

=> 3(x + 3)/(x - 3)(x + 3)

=> 3/(x - 3)

**The final simplified form is 3/(x - 3)**

We notice that the denominator of the 3rd fraction is a difference of squares that returns the product:

x^2 - 9 = (x-3)(x+3)

We notice that we can factorize by -1 the denominator of the 2nd fraction:

2/(3-x) = -2/(x-3)

We'll re-write theĀ sum of fractions:

3/(x+3) - 2/(x-3) + (2x+24)/ (x-3)(x+3)

We'll multiply each fraction by the missing factor in order to get the LCD (x-3)(x+3)

3(x-3)/(x-3)(x+3) - 2(x+3)/(x-3)(x+3) + (2x+24)/ (x-3)(x+3)

Since the fractions have the same denominators, we can re-write the sum:

[3(x-3) - 2(x+3) + (2x+24)]/ (x-3)(x+3)

(3x - 9 - 2x - 6 + 2x + 24)/ (x-3)(x+3)

We'll combine and eliminate like terms:

(3x - 9 - 6 + 24)/ (x-3)(x+3)

(3x+9)/ (x-3)(x+3)

We'll factorize the numerator by 3:

3(x+3)/ (x-3)(x+3)

We'll simplify and we'll get:

3/(x-3)

**The result of the given sum, reduced to the lowest terms, is 3/(x-3).**