What is the difference? \( \frac{2 x+5}{x^{2}-3 x}-\frac{3 x+5}{x^{3}-9 x}-\frac{x+1}{x^{2}-9} \)

To simplify the expression, let's first identify and factor the denominators. 1. The first denominator \( x^2 - 3x \) can be factored as \( x(x - 3) \). 2. The second denominator \( x^3 - 9x \) can be factored by taking out \( x \), giving us \( x(x^2 - 9) \). The expression \( x^2 - 9 \) is a difference of squares, which factors to \( x(x - 3)(x + 3) \). 3. The third denominator \( x^2 - 9 \) also factors as \( (x - 3)(x + 3) \). Now, let's write the common denominator. The least common denominator (LCD) will be \( x(x - 3)(x + 3) \). Now we rewrite each fraction with the common denominator: 1. For the first fraction: \[ \frac{2x + 5}{x^2 - 3x} = \frac{2x + 5}{x(x - 3)} \times \frac{(x + 3)}{(x + 3)} = \frac{(2x + 5)(x + 3)}{x(x - 3)(x + 3)} \] 2. For the second fraction: \[ \frac{3x + 5}{x^3 - 9x} = \frac{3x + 5}{x(x - 3)(x + 3)} \quad \text{(it is already in the appropriate form)} \] 3. For the third fraction: \[ \frac{x + 1}{x^2 - 9} = \frac{x + 1}{(x - 3)(x + 3)} \times \frac{x}{x} = \frac{x(x + 1)}{x(x - 3)(x + 3)} \] Now combine all three fractions: \[ \frac{(2x + 5)(x + 3)-(3x + 5) - x(x + 1)}{x(x - 3)(x + 3)} \] Next, simplify the numerator: 1. Expand \( (2x + 5)(x + 3) \): \[ = 2x^2 + 6x + 5x + 15 = 2x^2 + 11x + 15 \] Now combine everything in the numerator: \[ 2x^2 + 11x + 15 - (3x + 5) - (x^2 + x) \] Distributing the minus signs gives: \[ 2x^2 + 11x + 15 - 3x - 5 - x^2 - x \] Combining like terms: \[ (2x^2 - x^2) + (11x - 3x - x) + (15 - 5) = x^2 + 7x + 10 \] Now our expression is: \[ \frac{x^2 + 7x + 10}{x(x - 3)(x + 3)} \] Finally, we can factor the numerator: \[ x^2 + 7x + 10 = (x + 2)(x + 5) \] Thus, the final simplified form is: \[ \frac{(x + 2)(x + 5)}{x(x - 3)(x + 3)} \] Just remember, this expression is valid as long as \( x \neq 0 \), \( x \neq 3 \), and \( x \neq -3 \) to avoid division by zero.