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

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

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

We move all terms to the left:

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


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

n!=0/1

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

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


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


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

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

We add all the numbers together, and all the variables

20/n^2*n

We multiply all the terms by the denominator


20

Back to the equation:

+(20)


We add all the numbers together, and all the variables

n+(-1n)/n^2*n)-1+20=0

We multiply all the terms by the denominator


n*n^2*n)-1+20+(-1n)=0

Wy multiply elements

n^4*n-1n)=0

We do not support enpression: n^4