`Simplify(1/(2x+3y) )-((4x)/((4x^2)-(9y^2))) +(2/(2x-3y))`  

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`4x^2-9y^2 = (2x)^2-(3y)^2 = (2x+3y)(2x-3y)`

 

`1/(2x+3y)-(4x)/(4x^2-9y^2)+2/(2x-3y)`

`1/(2x+3y)-(4x)/((2x+3y)(2x-3y))+2/(2x-3y)`

 

Now we should get `(2x+3y)(2x-3y)` as common denominator.

 

`1/(2x+3y)-(4x)/((2x+3y)(2x-3y))+2/(2x-3y)`

`= (2x-3y)/((2x+3y)(2x-3y))-(4x)/((2x+3y)(2x-3y))+(2(2x+3y))/((2x+3y)(2x-3y))`

`= [(2x-3y)-4x+2(2x+3y)]/((2x+3y)(2x-3y))`

`= (2x+3y)/((2x+3y)(2x-3y))`

`= 1/(2x-3y)`

 

So the answer is

`1/(2x+3y)-(4x)/((2x+3y)(2x-3y))+2/(2x-3y) = 1/(2x-3y)`

 

 

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