Answer
$\frac{5x}{(x-6)(x-1)(x+4)}$.
Work Step by Step
The given expression is
$\Rightarrow \frac{x}{x^2-7x+6}-\frac{x}{x^2-2x-24}$
Factor each denominator as shown below.
$\Rightarrow x^2-7x+6$
Rewrite the middle term $-7x$ as $-6x-x$.
$\Rightarrow x^2-6x-x+6$
Group terms.
$\Rightarrow (x^2-6x)+(-x+6)$
Factor each group.
$\Rightarrow x(x-6)-1(x-6)$
Factor out $(x-6)$.
$\Rightarrow (x-6)(x-1)$
and
$\Rightarrow x^2-2x-24$
Rewrite the middle term $-2x$ as $-6x+4x$.
$\Rightarrow x^2-6x+4x-24$
Group terms.
$\Rightarrow (x^2-6x)+(4x-24)$
Factor each group.
$\Rightarrow x(x-6)+4(x-6)$
Factor out $(x-6)$.
$\Rightarrow (x-6)(x+4)$
Back substitute into the fractions.
$\Rightarrow \frac{x}{(x-6)(x-1)}-\frac{x}{(x-6)(x+4)}$
The LCD of the denominators is $(x-6)(x-1)(x+4)$.
$\Rightarrow \frac{x(x+4)}{(x-6)(x-1)(x+4)}-\frac{x(x-1)}{(x-6)(x-1)(x+4)}$
Add both numerators because denominators are equal.
$\Rightarrow \frac{x(x+4)-x(x-1)}{(x-6)(x-1)(x+4)}$
Use the distributive property.
$\Rightarrow \frac{x^2+4x-x^2+x}{(x-6)(x-1)(x+4)}$
Add like terms.
$\Rightarrow \frac{5x}{(x-6)(x-1)(x+4)}$.
