(12/x^2-16)-(24/x-4)=3

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Solution for (12/x^2-16)-(24/x-4)=3 equation:


D( x )

x = 0

x^2 = 0

x = 0

x = 0

x^2 = 0

x^2 = 0

1*x^2 = 0 // : 1

x^2 = 0

x = 0

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

12/(x^2)-(24/x)-16+4 = 3 // - 3

12/(x^2)-(24/x)-16-3+4 = 0

12/(x^2)-24*x^-1-16-3+4 = 0

12*x^-2-24*x^-1-15 = 0

t_1 = x^-1

12*t_1^2-24*t_1^1-15 = 0

12*t_1^2-24*t_1-15 = 0

DELTA = (-24)^2-(-15*4*12)

DELTA = 1296

DELTA > 0

t_1 = (1296^(1/2)+24)/(2*12) or t_1 = (24-1296^(1/2))/(2*12)

t_1 = 5/2 or t_1 = -1/2

t_1 = -1/2

x^-1+1/2 = 0

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

x^-1 = -1/2

-1 < 0

1/(x^1) = -1/2 // * x^1

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

-2 = x^1

x = -2

t_1 = 5/2

x^-1-5/2 = 0

1*x^-1 = 5/2 // : 1

x^-1 = 5/2

-1 < 0

1/(x^1) = 5/2 // * x^1

1 = 5/2*x^1 // : 5/2

2/5 = x^1

x = 2/5

x in { -2, 2/5 }

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