(x-2)/(2x)+1=(x+1)/(x)

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



(x-2)/(2x)+1=(x+1)/(x)

We move all terms to the left:

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



Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R





Domain of the equation: x)!=0

x!=0/1

x!=0

x∈R



We calculate fractions


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

We calculate fractions


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

We get rid of parentheses

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

We calculate fractions


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



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

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

We can not solve this equation

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