Question

Answer 1

\(A=\dfrac{PQ}{P^2-Q^2}\)

\(=\dfrac{4x^2y^2}{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}:\left[\dfrac{4x^2y^2}{\left(x^2-y^2\right)^2}-\dfrac{4x^2y^2}{\left(x^2+y^2\right)^2}\right]\)

\(=\dfrac{4x^2y^2}{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}:\left[4x^2y^2\left(\dfrac{1}{\left(x^2-y^2\right)^2}-\dfrac{1}{\left(x^2+y^2\right)^2}\right)\right]\)

\(=\dfrac{4x^2y^2}{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}:\left[4x^2y^2\cdot\dfrac{x^4+2x^2y^2+y^4-x^4+2x^2y^2-y^4}{\left(x-y\right)^2\left(x+y\right)^2\left(x^2+y^2\right)^2}\right]\)

\(=\dfrac{4x^2y^2}{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}\cdot\dfrac{\left(x-y\right)^2\left(x+y\right)^2\left(x^2+y^2\right)^2}{4x^2y^2\cdot4x^2y^2}\)

\(=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{4x^2y^2}\)