can someone help simplify((2x*y^-2*y^4)^-2) * ((4xy)/(2*x^-1*y-3))?

Answer 1

((2x*y^-2*y^4)^-2) * ((4xy)/(2*x^-1*y-3))

=(y^4/(4x^2y^8))*(2xy)(xy^3)

=1/2

Answer 2

When a term is raised to a negative exponent, it is the reciprocal; y^-2 = 1/(y^2); remember lowest common denominator when adding or subtracting fractions
1/[(2x/y^2) - 2y^4]^2 * (4xy)/[(2/x) - (1/y^3)]
1/[(2x - 2y^6)/y^2]^2 * (4xy)/[(2y^3 - x)/xy^3]
y^4/(4x^2 - 8xy^6 + 4y^12) * [(4xy)(xy^3)/(2y^3 - x)]
y^4/(4x^2 - 8xy^6 + 4y^12) * (4x^2 * y^4)/(2y^3 - x)
(4x^2 * y^8)/(8x^2 * y^3 - 16xy^9 + 8y^15 - 4x^3 + 8x^2 * y^6 - 4xy^12)
factoring out 4
(x^2 * y^8)/(2x^2 * y^3 - 4xy^9 + 2y^15 - x^3 + 2x^2 * y6 - xy^12)

Answer 3

= (2x^2y^2)^-2 * (4xy)/(2*x^-1*y-3)
= 1/(4x^4y^4) * 2x^2y^4
=1/(2x^2)