Answer
$\frac{5x^2-7x+4}{(x+2)(x-1)(x-2)}$.
Work Step by Step
First determine the Least Common denominator (LCD).
Factor $x^2-4$.
$\Rightarrow x^2-2^2$
Use special formula $A^2-B^2=(A+B)(A-B)$.
$\Rightarrow (x+2)(x-2)$
Factor $x^2+x-2$.
Rewrite the middle term $x$ as $2x-1x$.
$\Rightarrow x^2+2x-1x-2$
Group terms.
$\Rightarrow (x^2+2x)+(-1x-2)$
Factor each group.
$\Rightarrow x(x+2)-1(x+2)$
Factor out $(x+2)$.
$\Rightarrow (x+2)(x-1)$
Back substitute into the given expression.
$\Rightarrow \Rightarrow \frac{5x}{x^2-4}-\frac{2}{x^2+x-2}=\frac{5x}{(x+2)(x-2)}-\frac{2}{(x+2)(x-1)}$
The LCD is $(x+2)(x-1)(x-2)$.
Multiply the numerator and the denominator to form LCD at the denominators.
$\Rightarrow \frac{5x(x-1)}{(x+2)(x-1)(x-2)}-\frac{2(x-2)}{(x+2)(x-1)(x-2)}$
Add numerators because denominators are same.
$\Rightarrow \frac{5x(x-1)-2(x-2)}{(x+2)(x-1)(x-2)}$
Use the distributive property.
$\Rightarrow \frac{5x^2-5x-2x+4}{(x+2)(x-1)(x-2)}$
Add like terms.
$\Rightarrow \frac{5x^2-7x+4}{(x+2)(x-1)(x-2)}$.
