5+(25/(n-5))=125/(n^2-5n)

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Solution for 5+(25/(n-5))=125/(n^2-5n) equation: 


D( n )

n^2-(5*n) = 0

n-5 = 0

n^2-(5*n) = 0

n^2-(5*n) = 0

n^2-5*n = 0

n^2-5*n = 0

DELTA = (-5)^2-(0*1*4)

DELTA = 25

DELTA > 0

n = (25^(1/2)+5)/(1*2) or n = (5-25^(1/2))/(1*2)

n = 5 or n = 0

n-5 = 0

n-5 = 0

n-5 = 0 // + 5

n = 5

n in (-oo:0) U (0:5) U (5:+oo)

25/(n-5)+5 = 125/(n^2-(5*n)) // - 125/(n^2-(5*n))

25/(n-5)-(125/(n^2-(5*n)))+5 = 0

25/(n-5)-125*(n^2-5*n)^-1+5 = 0

25/(n-5)-125/(n^2-5*n)+5 = 0

n^2-5*n = 0

n^2-5*n = 0

n*(n-5) = 0

n-5 = 0 // + 5

n = 5

n*(n-5) = 0

25/(n-5)-125/(n*(n-5))+5 = 0

(25*n)/(n*(n-5))-125/(n*(n-5))+(5*n*(n-5))/(n*(n-5)) = 0

5*n*(n-5)+25*n-125 = 0

5*n^2+25*n-25*n-125 = 0

5*n^2-125 = 0

(5*n^2-125)/(n*(n-5)) = 0

(5*n^2-125)/(n*(n-5)) = 0 // * n*(n-5)

5*n^2-125 = 0

5*n^2 = 125 // : 5

n^2 = 25

n^2 = 25 // ^ 1/2

abs(n) = 5

n = 5 or n = -5

n in { 5} 

n = -5

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