(5x/x^2-9)=(1/3x+9)+(4/x-3)

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Solution for (5x/x^2-9)=(1/3x+9)+(4/x-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)

(5*x)/(x^2)-9 = (1/3)*x+4/x-3+9 // - (1/3)*x+4/x-3+9

(5*x)/(x^2)-((1/3)*x)-(4/x)-9-9+3 = 0

(-1/3)*x+(5*x)/(x^2)-4*x^-1-9-9+3 = 0

1*x^-1-1/3*x^1-15*x^0 = 0

(1*x^0-1/3*x^2-15*x^1)/(x^1) = 0 // * x^2

x^1*(1*x^0-1/3*x^2-15*x^1) = 0

x^1

(-1/3)*x^2-15*x+1 = 0

(-1/3)*x^2-15*x+1 = 0

DELTA = (-15)^2-(1*4*(-1/3))

DELTA = 679/3

DELTA > 0

x = ((679/3)^(1/2)+15)/(2*(-1/3)) or x = (15-(679/3)^(1/2))/(2*(-1/3))

x = -3/2*((679/3)^(1/2)+15) or x = -3/2*(15-(679/3)^(1/2))

x in { -3/2*((679/3)^(1/2)+15), -3/2*(15-(679/3)^(1/2))} 

x in { -3/2*((679/3)^(1/2)+15), -3/2*(15-(679/3)^(1/2)) }

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