3/(2x+1)-2/(2x-1)-(x+3)/(4x2+1)=0

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Solution for 3/(2x+1)-2/(2x-1)-(x+3)/(4x2+1)=0 equation: 



3/(2x+1)-2/(2x-1)-(x+3)/(4x^2+1)=0



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

We move all terms containing x to the left, all other terms to the right

2x!=-1

x!=-1/2

x!=-1/2

x∈R





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

We move all terms containing x to the left, all other terms to the right

2x!=1

x!=1/2

x!=1/2

x∈R





Domain of the equation: (4x^2+1)!=0

We move all terms containing x to the left, all other terms to the right

4x^2!=-1

x^2!=-1/4

x^2!=√-1/4

x!=NAN

x∈R



We calculate fractions


(3*(2x-1)*(4x^2+1))/((2x+1)*(2x-1)*(4x^2+1))+(-2*(2x+1)*(4x^2+1))/((2x+1)*(2x-1)*(4x^2+1))+(-(x+3)*(2x+1)*(2x-1))/((2x+1)*(2x-1)*(4x^2+1))=0



We calculate terms in parentheses: +(3*(2x-1)*(4x^2+1))/((2x+1)*(2x-1)*(4x^2+1)), so:

3*(2x-1)*(4x^2+1))/((2x+1)*(2x-1)*(4x^2+1)

We multiply all the terms by the denominator


3*(2x-1)*(4x^2+1))

Back to the equation:

+(3*(2x-1)*(4x^2+1)))





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

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

We multiply all the terms by the denominator


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

Back to the equation:

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





We calculate terms in parentheses: +(-(x+3)*(2x+1)*(2x-1))/((2x+1)*(2x-1)*(4x^2+1)), so:

-(x+3)*(2x+1)*(2x-1))/((2x+1)*(2x-1)*(4x^2+1)

We multiply all the terms by the denominator


-(x+3)*(2x+1)*(2x-1))

Back to the equation:

+(-(x+3)*(2x+1)*(2x-1)))



We add all the numbers together, and all the variables

(3*(2x-1)*(4x^2+1)))+(-2*(2x+1)*(4x^2+1)))+(-(x+3)*(2x+1)*(2x=0

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