Question

`5xy-4x^2-y^2` divided by `x+y` multiplied `y^2-x^2` divided `y^2-4xy`?

rational expression

Answer 1

`(5xy-4x^2-y^2)/(x+y) * (y^2-x^2)/(y^2-4xy) = -(x-y)^2/y`

with exclusions `x != -y` and `x != y/4`.

Explanation:

Given:

`(5xy-4x^2-y^2)/(x+y) * (y^2-x^2)/(y^2-4xy)`

Note that we can reverse the sign of both of the numerators to get:

`((4x^2-5xy+y^2)(x^2-y^2))/((x+y)(y^2-4xy)) = (((4x^2-xy)-(4xy-y^2))(x-y)color(red)(cancel(color(black)((x+y)))))/(color(red)(cancel(color(black)((x+y))))(y-4x)y)`

`color(white)(((4x^2-5xy+y^2)(x^2-y^2))/((x+y)(y^2-4xy))) = ((x(4x-y)-y(4x-y))(x-y))/((y-4x)y)`

`color(white)(((4x^2-5xy+y^2)(x^2-y^2))/((x+y)(y^2-4xy))) = ((x-y)(4x-y)(x-y))/((y-4x)y)`

`color(white)(((4x^2-5xy+y^2)(x^2-y^2))/((x+y)(y^2-4xy))) = -((x-y)color(red)(cancel(color(black)((4x-y))))(x-y))/(color(red)(cancel(color(black)((4x-y))))y)`

`color(white)(((4x^2-5xy+y^2)(x^2-y^2))/((x+y)(y^2-4xy))) = -(x-y)^2/y`

with exclusions `x != -y` and `x != y/4` due to the factors we have cancelled from the denominator.