Answer
$\frac{3x^2}{2x-5}$.
Work Step by Step
The given expression is
$\Rightarrow \frac{x^2-4x+4-y^2}{2x^2-11x+15}\cdot\frac{x^4y}{x-2+y}\div\frac{x^3y-2x^2y-x^2y^2}{3x-9}$
Factor each numerator and denominator as shown below.
$\Rightarrow x^2-4x+4-y^2$
$\Rightarrow x^2-2(x)(2)+2^2-y^2$
Use the perfect square trinomial formula
$A^2-2(A)(B)+B^2=(A-B)^2$.
$\Rightarrow (x-2)^2-y^2$
Use the special formula $A^2-B^2=(A+B)(A-B)$.
$\Rightarrow (x-2+y)(x-2-y)$.
$\Rightarrow 2x^2-11x+15$
Rewrite the middle term $-11x$ as $-6x-5x$.
$\Rightarrow 2x^2-6x-5x+15$
Group terms.
$\Rightarrow (2x^2-6x)+(-5x+15)$
Factor each group.
$\Rightarrow 2x(x-3)-5(x-3)$
Factor out $(x-3)$.
$\Rightarrow (x-3)(2x-5)$.
$\Rightarrow x^3y-2x^2y-x^2y^2$
Factor out $x^2y$ from each term.
$\Rightarrow x^2y(x-2-y)$.
$\Rightarrow 3x-9$
Factor out $3$ from each term.
$\Rightarrow 3(x-3)$.
Back substitute all values into the given expression.
$\Rightarrow \frac{(x-2+y)(x-2-y)}{(x-3)(2x-5)}\cdot\frac{x^4y}{x-2+y}\div\frac{x^2y(x-2-y)}{3(x-3)}$
Invert the divisor and multiply.
$\Rightarrow \frac{(x-2+y)(x-2-y)}{(x-3)(2x-5)}\cdot\frac{x^4y}{x-2+y}\cdot\frac{3(x-3)}{x^2y(x-2-y)}$
Cancel common terms.
$\Rightarrow \frac{3x^2}{2x-5}$.
