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(x-2)/(3x)+1=(x+2)/(x)
We move all terms to the left:
(x-2)/(3x)+1-((x+2)/(x))=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: x)!=0We calculate fractions
x!=0/1
x!=0
x∈R
(x-2)*x)/3x^2+(-((x+2)*3x)/3x^2+1=0
We calculate fractions
((x-2)*x)*3x^2)/(3x^2+(*3x^2)+(-((x+2)*3x)*3x^2)/(3x^2+(*3x^2)+1=0
We get rid of parentheses
((x-2)*x)*3x^2)/(3x^2+*3x^2+(-((x+2)*3x)*3x^2)/(3x^2+(*3x^2)+1=0
We calculate fractions
*3x^2+(((x-2)*x)*3x^2)*(3x^2+*3x^2+1)/((3x^2*(3x^2+(*3x^2)+1)+((-((x+2)*3x)*3x^2)*3x^2/((3x^2*(3x^2+(*3x^2)+1)=0
We calculate terms in parentheses: +(((x-2)*x)*3x^2)*(3x^2+*3x^2+1)/((3x^2*(3x^2+(*3x^2)+1)+((-((x+2)*3x)*3x^2)*3x^2/((3x^2*(3x^2+(*3x^2)+1), so:
((x-2)*x)*3x^2)*(3x^2+*3x^2+1)/((3x^2*(3x^2+(*3x^2)+1)+((-((x+2)*3x)*3x^2)*3x^2/((3x^2*(3x^2+(*3x^2)+1
We can not solve this equation
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