Answer
$\frac{1}{2y-x}$.
Work Step by Step
Factor each term of the given expression:
$\frac{x+2y}{x^2+4xy+4y^2}-\frac{2x}{x^2-4y^2}=\frac{x+2y}{(x+2y)(x+2y)}-\frac{2x}{(x+2y)(x-2y)}$
Cancel common terms.
$=\frac{1}{(x+2y)}-\frac{2x}{(x+2y)(x-2y)}$
LCD is $=(x+2y)(x-2y)$.
Multiply each numerator and denominator by the extra factor required to form the LCD.
$=\frac{1}{x+2y}\times \frac{x-2y}{x-2y}-\frac{2x}{(x+2y)(x-2y)}$
Simplify.
$=\frac{x-2y}{(x+2y)(x-2y)}-\frac{2x}{(x+2y)(x-2y)}$
Add the numerators.
$=\frac{x-2y-2x}{(x+2y)(x-2y)}$
Simplify.
$=\frac{-2y-x}{(x+2y)(x-2y)}$
Or we can write.
$=\frac{-(x+2y)}{(x+2y)(x-2y)}$
Cancel common terms.
$=-\frac{1}{x-2y}$.
Or we can write.
$=\frac{1}{2y-x}$.
