(5/x-2)-(20/x^2-4)=1

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Solution for (5/x-2)-(20/x^2-4)=1 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)

5/x-(20/(x^2))-2+4 = 1 // - 1

5/x-(20/(x^2))-2-1+4 = 0

5/x-20*x^-2-2-1+4 = 0

5*x^-1-20*x^-2+1 = 0

t_1 = x^-1

5*t_1^1-20*t_1^2+1 = 0

5*t_1-20*t_1^2+1 = 0

DELTA = 5^2-(-20*1*4)

DELTA = 105

DELTA > 0

t_1 = (105^(1/2)-5)/(-20*2) or t_1 = (-105^(1/2)-5)/(-20*2)

t_1 = (105^(1/2)-5)/(-40) or t_1 = (105^(1/2)+5)/40

t_1 = (105^(1/2)-5)/(-40)

x^-1-((105^(1/2)-5)/(-40)) = 0

1*x^-1 = (105^(1/2)-5)/(-40) // : 1

x^-1 = (105^(1/2)-5)/(-40)

-1 < 0

1/(x^1) = (105^(1/2)-5)/(-40) // * x^1

1 = ((105^(1/2)-5)/(-40))*x^1 // : (105^(1/2)-5)/(-40)

-40*(105^(1/2)-5)^-1 = x^1

x = -40*(105^(1/2)-5)^-1

t_1 = (105^(1/2)+5)/40

x^-1-((105^(1/2)+5)/40) = 0

1*x^-1 = (105^(1/2)+5)/40 // : 1

x^-1 = (105^(1/2)+5)/40

-1 < 0

1/(x^1) = (105^(1/2)+5)/40 // * x^1

1 = ((105^(1/2)+5)/40)*x^1 // : (105^(1/2)+5)/40

40*(105^(1/2)+5)^-1 = x^1

x = 40*(105^(1/2)+5)^-1

x in { -40*(105^(1/2)-5)^-1, 40*(105^(1/2)+5)^-1 }

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