(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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