(x+1)^2-1/(2x-1)^2+4x/2x-1=0

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Solution for (x+1)^2-1/(2x-1)^2+4x/2x-1=0 equation: 



(x+1)^2-1/(2x-1)^2+4x/2x-1=0



Domain of the equation: (2x-1)^2!=0

x∈R





Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R



We calculate fractions


(x+1)^2+(-2x)/((2x-1)^2*2x)+(4x*(2x-1)^2)/((2x-1)^2*2x)-1=0



We calculate terms in parentheses: +(-2x)/((2x-1)^2*2x), so:

-2x)/((2x-1)^2*2x

We multiply all the terms by the denominator


-2x)

Back to the equation:

+(-2x))





We calculate terms in parentheses: +(4x*(2x-1)^2)/((2x-1)^2*2x), so:

4x*(2x-1)^2)/((2x-1)^2*2x

We multiply all the terms by the denominator


4x*(2x-1)^2)

We multiply parentheses

8x^2-4x^2

We add all the numbers together, and all the variables

4x^2

Back to the equation:

+(4x^2)



determiningTheFunctionDomain
(-2x))+4x^2-1+(x+1)^2=0

We add all the numbers together, and all the variables

(-2x))+4x^2-1+(x=0

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