(2/x-3)-(4/x+3)=(8/x^2-9)

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Solution for (2/x-3)-(4/x+3)=(8/x^2-9) 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)

2/x-(4/x)-3-3 = 8/(x^2)-9 // - 8/(x^2)-9

2/x-(4/x)-(8/(x^2))-3-3+9 = 0

2/x-4*x^-1-8*x^-2-3-3+9 = 0

3-2*x^-1-8*x^-2 = 0

t_1 = x^-1

3-8*t_1^2-2*t_1^1 = 0

3-8*t_1^2-2*t_1 = 0

DELTA = (-2)^2-(-8*3*4)

DELTA = 100

DELTA > 0

t_1 = (100^(1/2)+2)/(-8*2) or t_1 = (2-100^(1/2))/(-8*2)

t_1 = -3/4 or t_1 = 1/2

t_1 = -3/4

x^-1+3/4 = 0

1*x^-1 = -3/4 // : 1

x^-1 = -3/4

-1 < 0

1/(x^1) = -3/4 // * x^1

1 = -3/4*x^1 // : -3/4

-4/3 = x^1

x = -4/3

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

x in { -4/3, 2 }

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