Answer
$\dfrac{1}{x+2}+\dfrac{1}{x^{2}-4}-\dfrac{2}{x^{2}-x-2}=\dfrac{x^{2}-2x-5}{(x+2)(x-2)(x+1)}$
Work Step by Step
$\dfrac{1}{x+2}+\dfrac{1}{x^{2}-4}-\dfrac{2}{x^{2}-x-2}$
Factor the denominators of the second and third terms:
$\dfrac{1}{x+2}+\dfrac{1}{x^{2}-4}-\dfrac{2}{x^{2}-x-2}=...$
$...=\dfrac{1}{x+2}+\dfrac{1}{(x-2)(x+2)}-\dfrac{2}{(x-2)(x+1)}=...$
Evaluate the operations indicated using the LCD (Least common denominator) of the rational expressions and simplify if possible:
$...=\dfrac{(x-2)(x+1)+(x+1)-2(x+2)}{(x+2)(x-2)(x+1)}=...$
$...=\dfrac{x^{2}-x-2+x+1-2x-4}{(x+2)(x-2)(x+1)}=\dfrac{x^{2}-2x-5}{(x+2)(x-2)(x+1)}$