Question

what is the difference?
\\(\frac{2x + 5}{x^2 - 3x} - \frac{3x + 5}{x^3 - 9x} - \frac{x + 1}{x^2 - 9}\\)
\\(\circ \frac{(x + 5)(x + 2)}{x^3 - 9x}\\)\\(\circ \frac{(x + 5)(x + 4)}{x^3 - 9x}\\)\\(\circ \frac{-2x + 11}{x^3 - 12x - 9}\\)\\(\circ \frac{3(x + 2)}{x^2 - 3x}\\)

Explanation:

Step1: Factor all denominators

1. $x^2-3x = x(x-3)$
2. $x^3-9x = x(x^2-9) = x(x-3)(x+3)$
3. $x^2-9 = (x-3)(x+3)$

The least common denominator (LCD) is $x(x-3)(x+3) = x^3-9x$.

Step2: Rewrite fractions with LCD

1. $\frac{2x+5}{x(x-3)} = \frac{(2x+5)(x+3)}{x(x-3)(x+3)} = \frac{(2x+5)(x+3)}{x^3-9x}$
2. $\frac{3x+5}{x^3-9x}$ (already has LCD)
3. $\frac{x+1}{(x-3)(x+3)} = \frac{(x+1)x}{x(x-3)(x+3)} = \frac{x(x+1)}{x^3-9x}$

Step3: Expand numerators

1. $(2x+5)(x+3) = 2x^2 + 6x + 5x + 15 = 2x^2 + 11x + 15$
2. $x(x+1) = x^2 + x$

Step4: Combine fractions and simplify

$$\begin{align*} &\frac{2x^2 + 11x + 15 - (3x+5) - (x^2 + x)}{x^3-9x}\\ =&\frac{2x^2 + 11x + 15 - 3x - 5 - x^2 - x}{x^3-9x}\\ =&\frac{(2x^2 - x^2) + (11x - 3x - x) + (15 - 5)}{x^3-9x}\\ =&\frac{x^2 + 7x + 10}{x^3-9x} \end{align*}$$

Step5: Factor the final numerator

$x^2 + 7x + 10 = (x+5)(x+2)$

Answer:

$\boldsymbol{\frac{(x+5)(x+2)}{x^3-9x}}$ (first option)