Answer
$\frac{1}{x-2y}$.
Work Step by Step
The given expression is
$\frac{x^2-4y^2}{x^2+3xy+2y^2}\div \frac{x^2-4xy+4y^2}{x+y}$
Replace divide sign with multiplication sign and reciprocal the right hand side fraction as shown below.
$\frac{x^2-4y^2}{x^2+3xy+2y^2}\times \frac{x+y}{x^2-4xy+4y^2}$
Factor each denominator and numerator.
$=x^2-4y^2$
Use algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=x^2-(2y)^2$
$=(x+2y)(x-2y)$
$=x^2+3xy+2y^2$
Rewrite middle term.
$=x^2+2xy+xy+2y^2$
Group terms.
$=(x^2+2xy)+(xy+2y^2)$
Factor from each group.
$=x(x+2y)+y(x+2y)$
Factor out common term.
$=(x+2y)(x+y)$
$=x+y$
This is the prime polynomial.
$=x^2-4xy+4y^2$
Rewrite middle term.
$=x^2-2xy-2xy+4y^2$
Group terms.
$=(x^2-2xy)+(-2xy+4y^2)$
Factor from each group.
$=x(x-2y)-2y(x-2y)$
Factor out common term.
$=(x-2y)(x-2y)$
Substitute all factors into the given expression.
$\frac{(x+2y)(x-2y)}{(x+2y)(x+y)}\times \frac{x+y}{(x-2y)(x-2y)}$
Cancel common factors.
The remaining fraction is
$=\frac{1}{x-2y}$.
