((n^2-4n)/(n-1))+(5/(n-1))=n+1

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Solution for ((n^2-4n)/(n-1))+(5/(n-1))=n+1 equation: 


D( n )

n-1 = 0

n-1 = 0

n-1 = 0

n-1 = 0 // + 1

n = 1

n in (-oo:1) U (1:+oo)

(n^2-(4*n))/(n-1)+5/(n-1) = n+1 // - n+1

(n^2-(4*n))/(n-1)+5/(n-1)-n-1 = 0

(n^2-4*n)/(n-1)+5/(n-1)-n-1 = 0

(n^2-4*n)/(n-1)+5/(n-1)+(-n*(n-1))/(n-1)+(-1*(n-1))/(n-1) = 0

n^2-n*(n-1)-1*(n-1)-4*n+5 = 0

n^2-n^2-4*n+n-n+1+5 = 0

1-3*n-n+5 = 0

6-4*n = 0

(6-4*n)/(n-1) = 0

(6-4*n)/(n-1) = 0 // * n-1

6-4*n = 0

6-4*n = 0 // - 6

-4*n = -6 // : -4

n = -6/(-4)

n = 3/2

n = 3/2

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