25/m-m/5+m

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Solution for 25/m-m/5+m equation: 


D( m )

m = 0

m = 0

m = 0

m in (-oo:0) U (0:+oo)

m-(m/5)+25/m = 0

4/5*m^1+25*m^-1 = 0

(4/5*m^2+25*m^0)/(m^1) = 0 // * m^2

m^1*(4/5*m^2+25*m^0) = 0

m^1

(4/5)*m^2+25 = 0

(4/5)*m^2+25 = 0

DELTA = 0^2-(4*25*(4/5))

DELTA = -80

DELTA < 0

m in { } 

m belongs to the empty set

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