Answer
$\Big(\dfrac{x^{2}-9}{x^{2}-1}\cdot\dfrac{x^{2}+2x+1}{2x^{2}+9x+9}\Big)\div\dfrac{2x+3}{1-x}=-\dfrac{(x-3)(x+1)}{(2x+3)^{2}}$
Work Step by Step
$\Big(\dfrac{x^{2}-9}{x^{2}-1}\cdot\dfrac{x^{2}+2x+1}{2x^{2}+9x+9}\Big)\div\dfrac{2x+3}{1-x}$
Factor the expressions inside the parentheses completely:
$\Big(\dfrac{(x-3)(x+3)}{(x-1)(x+1)}\cdot\dfrac{(x+1)^{2}}{(x+3)(2x+3)}\Big)\div\dfrac{2x+3}{1-x}=...$
Evaluate the product of the two rational expressions inside the parentheses:
$...=\dfrac{(x-3)(x+3)(x+1)^{2}}{(x-1)(x+1)(x+3)(2x+3)}\div\dfrac{2x+3}{1-x}=...$
Evaluate the division:
$...=\dfrac{(x-3)(x+3)(x+1)^{2}(1-x)}{(x-1)(x+1)(x+3)(2x+3)^{2}}=...$
Simplify by removing the factors that appear both in the numerator and the denominator. To eliminate $(1-x)$ and $(x-1)$, change the sign of the first factor and also change the sign of the fraction:
$...=-\dfrac{(x-3)(x+3)(x+1)^{2}(x-1)}{(x-1)(x+1)(x+3)(2x+3)^{2}}=-\dfrac{(x-3)(x+1)}{(2x+3)^{2}}$
