(16x^2-48)/(39(x^2-1)^(4/3))=0

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Solution for (16x^2-48)/(39(x^2-1)^(4/3))=0 equation: 


D( x )

39*(x^2-1)^(4/3) = 0

39*(x^2-1)^(4/3) = 0

39*(x^2-1)^(4/3) = 0

( 39 )

39 = 0

x belongs to the empty set

( x^2-1 )

1*x^2 = 1 // : 1

x^2 = 1

x^2 = 1 // ^ 1/2

abs(x) = 1

x = 1 or x = -1

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

(16*x^2-48)/(39*(x^2-1)^(4/3)) = 0

16*x^2 = 48 // : 16

x^2 = 3

x^2 = 3 // ^ 1/2

abs(x) = 3^(1/2)

x = 3^(1/2) or x = -3^(1/2)

x in { 3^(1/2), -3^(1/2) }

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