t^(5/3)-25t^(-1/3)=0

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Solution for t^(5/3)-25t^(-1/3)=0 equation:


D( t )

t = 0

t = 0

t = 0

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

t^(5/3)-(25*t^(-1/3)) = 0

t^(5/3)-25*t^(-1/3) = 0

t_1 = t^(1/3)

1*t_1^5-25*t_1^-1 = 0

t_1^5-25*t_1^-1 = 0

t_1^-1*(t_1^6-25) = 0

1*t_1^6 = 25 // : 1

t_1^6 = 25

t_1^6 = 25 // ^ 1/6

abs(t_1) = 25^(1/6)

t_1 = 25^(1/6) or t_1 = -25^(1/6)

1/t_1 = 0

1*t_1^-1 = 0 // : 1

t_1^-1 = 0

t_1 należy do O

t_1 = 25^(1/6)

t^(1/3)-25^(1/6) = 0

1*t^(1/3) = 25^(1/6) // : 1

t^(1/3) = 25^(1/6)

t^(1/3) = 25^(1/6) // ^ 3

t = 5

t_1 = -25^(1/6)

t^(1/3)+25^(1/6) = 0

1*t^(1/3) = -25^(1/6) // : 1

t^(1/3) = -25^(1/6)

( -25^(1/6) < 0 i 1/3 in (0:1) ) => t należy do O

t = 5

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