Answer
$\frac{3x^2+9x}{x^2+8x-33}$.
Work Step by Step
The given expression is
$\Rightarrow \frac{\frac{2}{x^2-x-6}+\frac{1}{x^2-4x+3}}{\frac{3}{x^2+x-2}-\frac{2}{x^2+5x+6}}$
Factor each denominator as shown below.
$\Rightarrow x^2-x-6$
Rewrite the middle term $-x$ as $-3x+2x$.
$\Rightarrow x^2-3x+2x-6$
Group terms.
$\Rightarrow (x^2-3x)+(2x-6)$
Factor each group.
$\Rightarrow x(x-3)+2(x-3)$
Factor out $(x-3)$.
$\Rightarrow (x-3)(x+2)$
$\Rightarrow x^2+x-2$
Rewrite the middle term $x$ as $2x-x$.
$\Rightarrow x^2+2x-x-2$
Group terms.
$\Rightarrow (x^2+2x)+(-x-2)$
Factor each group.
$\Rightarrow x(x+2)-1(x+1)$
Factor out $(x+2)$.
$\Rightarrow (x+2)(x-1)$
$\Rightarrow x^2-4x+3$
Rewrite the middle term $-4x$ as $-3x-x$.
$\Rightarrow x^2-3x-x+3$
Group terms.
$\Rightarrow (x^2-3x)+(-x+3)$
Factor each group.
$\Rightarrow x(x-3)-1(x-3)$
Factor out $(x-3)$.
$\Rightarrow (x-3)(x-1)$
$\Rightarrow x^2+5x+6$
Rewrite the middle term $5x$ as $3x+2x$.
$\Rightarrow x^2+3x+2x+6$
Group terms.
$\Rightarrow (x^2+3x)+(2x+6)$
Factor each group.
$\Rightarrow x(x+3)+2(x+3)$
Factor out $(x+3)$.
$\Rightarrow (x+3)(x+2)$
Back substitute all factors into the given expression.
$\Rightarrow \frac{\frac{2}{(x-3)(x+2)}+\frac{1}{(x-3)(x-1)}}{\frac{3}{(x+2)(x-1)}-\frac{2}{(x+3)(x+2)}}$
The LCD of all the denominators is $(x-3)(x-1)(x+2)(x+3)$
Multiply the numerator and the denominators by $(x-3)(x-1)(x+2)(x+3)$.
$\Rightarrow \frac{(x-3)(x-1)(x+2)(x+3)\left (\frac{2}{(x-3)(x+2)}+\frac{1}{(x-3)(x-1)}\right)}{(x-3)(x-1)(x+2)(x+3)\left (\frac{3}{(x+2)(x-1)}-\frac{2}{(x+3)(x+2)}\right )}$
$\Rightarrow \frac{2(x-1)(x+3)+1(x+2)(x+3)}{3(x-3)(x+3)-2(x-3)(x-1)}$
Use the FOIL method.
$\Rightarrow \frac{2(x^2+3x-x-3)+1(x^2+3x+2x+6)}{3(x^2+3x-3x-9)-2(x^2-x-3x+3)}$
Simplify.
$\Rightarrow \frac{2(x^2+2x-3)+1(x^2+5x+6)}{3(x^2-9)-2(x^2-4x+3)}$
Use the distributive property.
$\Rightarrow \frac{2x^2+4x-6+x^2+5x+6}{3x^2-27-2x^2+8x-6}$
Add like terms.
$\Rightarrow \frac{3x^2+9x}{x^2+8x-33}$.
