(n+2)^2-(n+2)^2/2n^2+3n=1/n

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Solution for (n+2)^2-(n+2)^2/2n^2+3n=1/n equation: 



(n+2)^2-(n+2)^2/2n^2+3n=1/n

We move all terms to the left:

(n+2)^2-(n+2)^2/2n^2+3n-(1/n)=0



Domain of the equation: 2n^2!=0

n^2!=0/2

n^2!=√0

n!=0

n∈R





Domain of the equation: n)!=0

n!=0/1

n!=0

n∈R



We add all the numbers together, and all the variables

(n+2)^2-(n+2)^2/2n^2+3n-(+1/n)=0

We add all the numbers together, and all the variables

3n+(n+2)^2-(n+2)^2/2n^2-(+1/n)=0

We get rid of parentheses

3n+(n+2)^2-(n+2)^2/2n^2-1/n=0

We calculate fractions

We do not support enpression: n^3

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