((4x+40)/(x^2-100))

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Solution for ((4x+40)/(x^2-100)) equation: 


D( x )

x^2-100 = 0

x^2-100 = 0

x^2-100 = 0

1*x^2 = 100 // : 1

x^2 = 100

x^2 = 100 // ^ 1/2

abs(x) = 10

x = 10 or x = -10

x in (-oo:-10) U (-10:10) U (10:+oo)

(4*x+40)/(x^2-100) = 0

4*x+40 = 0 // - 40

4*x = -40 // : 4

x = -40/4

x = -10

x in { -10} 

x belongs to the empty set

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