((x^2+29)/(x^2-4x-5))-((x+4)/(x-5))

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Solution for ((x^2+29)/(x^2-4x-5))-((x+4)/(x-5)) equation: 


D( x )

x-5 = 0

x^2-(4*x)-5 = 0

x-5 = 0

x-5 = 0

x-5 = 0 // + 5

x = 5

x^2-(4*x)-5 = 0

x^2-(4*x)-5 = 0

x^2-4*x-5 = 0

x^2-4*x-5 = 0

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

DELTA = 36

DELTA > 0

x = (36^(1/2)+4)/(1*2) or x = (4-36^(1/2))/(1*2)

x = 5 or x = -1

x in (-oo:-1) U (-1:5) U (5:+oo)

(x^2+29)/(x^2-(4*x)-5)-((x+4)/(x-5)) = 0

(x^2+29)/(x^2-4*x-5)-((x+4)/(x-5)) = 0

(x^2+29)/(x^2-4*x-5)+(-1*(x+4))/(x-5) = 0

x^2-4*x-5 = 0

x^2-4*x-5 = 0

x^2-4*x-5 = 0

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

DELTA = 36

DELTA > 0

x = (36^(1/2)+4)/(1*2) or x = (4-36^(1/2))/(1*2)

x = 5 or x = -1

(x+1)*(x-5) = 0

(x^2+29)/((x+1)*(x-5))+(-1*(x+4))/(x-5) = 0

(x^2+29)/((x+1)*(x-5))+(-1*(x+4)*(x+1))/((x+1)*(x-5)) = 0

x^2-1*(x+4)*(x+1)+29 = 0

25-5*x = 0

(25-5*x)/((x+1)*(x-5)) = 0

(25-5*x)/((x+1)*(x-5)) = 0 // * (x+1)*(x-5)

25-5*x = 0

25-5*x = 0 // - 25

-5*x = -25 // : -5

x = -25/(-5)

x = 5

x in { 5} 

x belongs to the empty set

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