(5/t-3)-(30/t^2-9)=1

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Solution for (5/t-3)-(30/t^2-9)=1 equation:


D( t )

t^2 = 0

t = 0

t^2 = 0

t^2 = 0

1*t^2 = 0 // : 1

t^2 = 0

t = 0

t = 0

t = 0

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

5/t-(30/(t^2))-3+9 = 1 // - 1

5/t-(30/(t^2))-3-1+9 = 0

5/t-30*t^-2-3-1+9 = 0

5*t^-1-30*t^-2+5 = 0

t_1 = t^-1

5*t_1^1-30*t_1^2+5 = 0

5*t_1-30*t_1^2+5 = 0

DELTA = 5^2-(-30*4*5)

DELTA = 625

DELTA > 0

t_1 = (625^(1/2)-5)/(-30*2) or t_1 = (-625^(1/2)-5)/(-30*2)

t_1 = -1/3 or t_1 = 1/2

t_1 = -1/3

t^-1+1/3 = 0

1*t^-1 = -1/3 // : 1

t^-1 = -1/3

-1 < 0

1/(t^1) = -1/3 // * t^1

1 = -1/3*t^1 // : -1/3

-3 = t^1

t = -3

t_1 = 1/2

t^-1-1/2 = 0

1*t^-1 = 1/2 // : 1

t^-1 = 1/2

-1 < 0

1/(t^1) = 1/2 // * t^1

1 = 1/2*t^1 // : 1/2

2 = t^1

t = 2

t in { -3, 2 }

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