Answer
$\frac{x+4}{x(x+2)}$.
Work Step by Step
The given expression is
$=\frac{1}{x}+\frac{4}{x^2-4}-\frac{2}{x^2-2x}$
First denominator $=x$.
Second denominator $=x^2-4$.
Use the algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=x^2-(2)^2$
$=(x+2)(x-2)$
Third denominator $=x^2-2x$.
Factor $=x(x-2)$.
Substitute all factors into the given expression.
$=\frac{1}{x}+\frac{4}{(x+2)(x-2)}-\frac{2}{x(x-2)}$
The LCM of all the denominators is
$=x(x+2)(x-2)$.
Multiply all fractions to make denominators equal.
$=\frac{1}{x}\times \frac{(x+2)(x-2)}{(x+2)(x-2)}+\frac{4}{(x+2)(x-2)}\times \frac{x}{x}-\frac{2}{x(x-2)}\times \frac{(x+2)}{(x+2)}$
Simplify.
$=\frac{(x+2)(x-2)}{x(x+2)(x-2)}+\frac{4x}{x(x+2)(x-2)}-\frac{2(x+2)}{x(x+2)(x-2)}$
$=\frac{(x+2)(x-2)+4x-2(x+2)}{x(x+2)(x-2)}$
$=\frac{x^2-4+4x-2x-4}{x(x+2)(x-2)}$
Simplify.
$=\frac{x^2+2x-8}{x(x+2)(x-2)}$
Factor the numerator
$=x^2+2x-8$
$=x^2+4x-2x-8$
$=(x^2+4x)+(-2x-8)$
$=x(x+4)-2(x+4)$
$=(x+4)(x-2)$
Substitute the factor.
$=\frac{(x+4)(x-2)}{x(x+2)(x-2)}$
Cancel common terms.
$=\frac{x+4}{x(x+2)}$.
