[x/(x-y)] - [2x/(x+y)] + [2xy/(y^2-x^2)]?

Question

Simplify.

Answer 1

let see first you want have all the like denominators

x(x+y) - 2x(x-y) - 2xy
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(x-y)(x+y)

just remember pull out a -1 and change the third part of the equation to x^2-y^2

x^2 + xy - 2x^2 +2xy -2xy / (x-y)(x+y)
xy - x^2 / (x+y)(x-y)
-x(x-y)/(x+y)(x-y)
-x/(x+y)

Answer 2

= 5

Answer 3

okay.[x/(x-y)] - [ 2x/(x+y)] equals [-x/(x+y)]. Then you add the [2xy/(y^2 - x^2)]. To do this, you must have a common denominator so multiply the top and bottom of [-x/(x+y)] by (y^2-x^2) and you'll get [(-x(y^2) + x ^3)/((x+y)(y^2-x^2))]. Then multiply the other equation by (x+y), top and bottom to get [(2(x^2)y - 2x(y^2))/((x+y)(y^2 - x^2))]. Then you just add the numerators to get
[(2(x^2)y - 3xy^2 +x^3)/((x+y)(y^2-x^2))]

Answer 4

[x/(x-y)] - [2x/(x+y)] + [2xy/(y^2-x^2)]
=[x/(x-y)] - [2x/(x+y)] + [-2xy/(x^2-y^2)]
= x(x+y)/(x^2-y^2) -2x(x-y)/(x^+y^2) -2xy/(x^2-y^2)]
= (x^2+xy -2x^2+2xy -2xy)/(x^2-y^2)
= (-x^2 +xy)/(x^2-y^2)
= -x(x-y)/(x^2-y^2)]
= -x/(x+y)

Answer 5

(x(x+y)-2x(x-y)-2xy)/(x^2-y^2)=

(x^2 +xy -2x^2 +2xy -2xy)/(x^2-y^2)=

(xy-y^2)/(x^2-y^2)=

y(x-y)/(x+y)(x-y)=

y/x+y=

Answer 6

[x/(x-y)] -[2x/(x+y)] +[2xy/(y^2-x^2)]
=[(x(x+y)-2x(x-y)]/(x+y)(x-y) + [2xy/ (x+y)(y-x)]
=[x^2+xy-2x^2+2xy]/(x+y)(x-y) +[-2xy / (x+y)(x-y)]
=[3xy-x^2](x+y)(x-y) -[2xy]/(x+y)(x-y)
=[xy-x^2]/(x+y)(x-y)
=x(y-x) / (x+y)(x-y)
=-x(x-y)/(x+y)(x-y)
=-x/(x+y)
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Answer 7

that's easy!!!!!!!!!!!
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