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(1/(n+3))+(5/(n^2-9))=2/(n-3)
We move all terms to the left:
(1/(n+3))+(5/(n^2-9))-(2/(n-3))=0
Domain of the equation: (n+3))!=0
n∈R
Domain of the equation: (n^2-9))!=0
n∈R
Domain of the equation: (n-3))!=0We calculate fractions
n∈R
((1*(n^2-9))*n)/((n+3))*(n^2-9))*(n-3)))+((5*(n+3))*n)/((n+3))*(n^2-9))*(n-3)))+(-(2*(n+3))*n^2)/((n+3))*(n^2-9))*(n-3)))=0
We can not solve this equation
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