$$ \frac{x}{x^{2}-2 x-24}-\frac{x}{x^{2}-8 x+12}=\square $$ (Simplify your answer.)

Question

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Factor the expression by following steps:
- step0: Factor:
  $$x^{2}-8x+12$$
- step1: Rewrite the expression:
  $$x^{2}+\left(-2-6\right)x+12$$
- step2: Rewrite the expression:
  $$x^{2}-2x-6x+12$$
- step3: Factor the expression:
  $$x\left(x-2\right)-6\left(x-2\right)$$
- step4: Factor the expression:
  $$\left(x-6\right)\left(x-2\right)$$
Factor the expression $$ x^{2}-2 x-24 $$.
Factor the expression by following steps:
- step0: Factor:
  $$x^{2}-2x-24$$
- step1: Rewrite the expression:
  $$x^{2}+\left(4-6\right)x-24$$
- step2: Rewrite the expression:
  $$x^{2}+4x-6x-24$$
- step3: Factor the expression:
  $$x\left(x+4\right)-6\left(x+4\right)$$
- step4: Factor the expression:
  $$\left(x-6\right)\left(x+4\right)$$
Calculate or simplify the expression $$ \frac{x}{(x-6)(x+4)}-\frac{x}{(x-6)(x-2)} $$.
Simplify the expression by following steps:
- step0: Solution:
  $$\frac{x}{\left(x-6\right)\left(x+4\right)}-\frac{x}{\left(x-6\right)\left(x-2\right)}$$
- step1: Reduce fractions to a common denominator:
  $$\frac{x\left(x-2\right)}{\left(x-6\right)\left(x+4\right)\left(x-2\right)}-\frac{x\left(x+4\right)}{\left(x-6\right)\left(x-2\right)\left(x+4\right)}$$
- step2: Rewrite the expression:
  $$\frac{x\left(x-2\right)}{\left(x-6\right)\left(x+4\right)\left(x-2\right)}-\frac{x\left(x+4\right)}{\left(x-6\right)\left(x+4\right)\left(x-2\right)}$$
- step3: Transform the expression:
  $$\frac{x\left(x-2\right)-x\left(x+4\right)}{\left(x-6\right)\left(x+4\right)\left(x-2\right)}$$
- step4: Multiply the terms:
  $$\frac{x^{2}-2x-x\left(x+4\right)}{\left(x-6\right)\left(x+4\right)\left(x-2\right)}$$
- step5: Multiply the terms:
  $$\frac{x^{2}-2x-\left(x^{2}+4x\right)}{\left(x-6\right)\left(x+4\right)\left(x-2\right)}$$
- step6: Subtract the terms:
  $$\frac{-6x}{\left(x-6\right)\left(x+4\right)\left(x-2\right)}$$
- step7: Rewrite the fraction:
  $$-\frac{6x}{\left(x-6\right)\left(x+4\right)\left(x-2\right)}$$
- step8: Calculate:
  $$-\frac{6x}{x^{3}-4x^{2}-20x+48}$$
To simplify the expression

$$
\frac{x}{x^{2}-2x-24}-\frac{x}{x^{2}-8x+12}
$$

we first need to factor the denominators.

1. **Factoring the first denominator**:
   $$
   x^{2}-2x-24 = (x-6)(x+4)
   $$

2. **Factoring the second denominator**:
   $$
   x^{2}-8x+12 = (x-6)(x-2)
   $$

Now we can rewrite the expression with the factored denominators:

$$
\frac{x}{(x-6)(x+4)} - \frac{x}{(x-6)(x-2)}
$$

Next, we find a common denominator, which is $$(x-6)(x+4)(x-2)$$. We rewrite each fraction:

$$
\frac{x(x-2)}{(x-6)(x+4)(x-2)} - \frac{x(x+4)}{(x-6)(x-2)(x+4)}
$$

Now, we can combine the fractions:

$$
\frac{x(x-2) - x(x+4)}{(x-6)(x+4)(x-2)}
$$

Simplifying the numerator:

$$
x(x-2) - x(x+4) = x^2 - 2x - x^2 - 4x = -6x
$$

Thus, we have:

$$
\frac{-6x}{(x-6)(x+4)(x-2)}
$$

Finally, we can express the simplified result as:

$$
-\frac{6x}{x^{3}-4x^{2}-20x+48}
$$

This is the simplified form of the original expression.
$$ -\frac{6x}{x^{3}-4x^{2}-20x+48} $$

(Simplify your answer.)

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