(X+5)/abs(x+5)

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Solution for (X+5)/abs(x+5) equation: 


D( x )

abs(x+5) = 0

abs(x+5) = 0

abs(x+5) = 0

abs(x+5) = 0

x in (-oo:+oo)

abs(x+5)

x+5 >= 0

x+5 >= 0 // - 5

x >= -5

x in <-5:+oo)

x+5 = 0

x in (-oo:-5)

-(x+5) = 0

abs(x+5) = 0

/| -(x+5) = 0 i x in (-oo:-5)| x+5 = 0 i x in <-5:+oo)

x in (-oo:-5)

-(x+5) = 0

-x-5 = 0 // + 5

-x = 5 // * -1

x = -5

(-oo:-5)

x in <-5:+oo)

x+5 = 0 // - 5

x = -5

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

(X+5)/abs(x+5) = 0

(X+5)/abs(x+5) = 0

x in (-oo:+oo)

abs(x+5)

x+5 >= 0

x+5 >= 0 // - 5

x >= -5

x in <-5:+oo)

(X+5)/(x+5) = 0

x in (-oo:-5)

(X+5)/(-(x+5)) = 0

(X+5)/abs(x+5) = 0

/| (X+5)/(-(x+5)) = 0 i x in (-oo:-5)| (X+5)/(x+5) = 0 i x in <-5:+oo)

x in (-oo:-5)

(X+5)/(-(x+5)) = 0

(X+5)/(-x-5) = 0

x in <-5:+oo)

(X+5)/(x+5) = 0

x belongs to the empty set

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