k+1/k=1-(((k^2)-3k-4)/4k)

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Solution for k+1/k=1-(((k^2)-3k-4)/4k) equation: 



k+1/k=1-(((k^2)-3k-4)/4k)

We move all terms to the left:

k+1/k-(1-(((k^2)-3k-4)/4k))=0



Domain of the equation: k!=0

k∈R





Domain of the equation: 4k))!=0

k!=0/1

k!=0

k∈R



We calculate fractions


k+4k/4k^2+(-(1-((k^2-3k-4)*k)/4k^2=0

We multiply all the terms by the denominator


k*4k^2+4k+(-(1-((k^2-3k-4)*k)=0



We calculate terms in parentheses: +(-(1-((k^2-3k-4)*k), so:

-(1-((k^2-3k-4)*k



We add all the numbers together, and all the variables

4k+k*4k^2+(-(1-((k^2-3k-4)*k)=0

Wy multiply elements

4k^3+4k+(-(1-((k^2-3k-4)*k)=0



We calculate terms in parentheses: +(-(1-((k^2-3k-4)*k), so:

-(1-((k^2-3k-4)*k



We do not support ekpression: k^3

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