(3/x-8)+(2/x-5)=(4/x^2-13x+40)

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Solution for (3/x-8)+(2/x-5)=(4/x^2-13x+40) 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)

3/x+2/x-8-5 = 4/(x^2)-(13*x)+40 // - 4/(x^2)-(13*x)+40

13*x+3/x+2/x-(4/(x^2))-40-8-5 = 0

13*x+3/x+2/x-4*x^-2-40-8-5 = 0

13*x^1+5*x^-1-4*x^-2-53*x^0 = 0

(13*x^3-53*x^2+5*x^1-4*x^0)/(x^2) = 0 // * x^4

x^2*(13*x^3-53*x^2+5*x^1-4*x^0) = 0

x^2

13*x^3-53*x^2+5*x-4 = 0

{ 1, -1, 2, -2, 4, -4 }

1

x = 1

13*x^3-53*x^2+5*x-4 = -39

1

-1

x = -1

13*x^3-53*x^2+5*x-4 = -75

-1

2

x = 2

13*x^3-53*x^2+5*x-4 = -102

2

-2

x = -2

13*x^3-53*x^2+5*x-4 = -330

-2

4

x = 4

13*x^3-53*x^2+5*x-4 = 0

4

x-4

13*x^2-x+1

13*x^3-53*x^2+5*x-4

x-4

52*x^2-13*x^3

5*x-x^2-4

x^2-4*x

x-4

4-x

0

13*x^2-x+1 = 0

DELTA = (-1)^2-(1*4*13)

DELTA = -51

DELTA < 0

x in { 4} 

x = 4

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