(3/3x+3)+(2/x-1)=2

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Solution for (3/3x+3)+(2/x-1)=2 equation: 


D( x )

x = 0

x = 0

x = 0

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

(3/3)*x+2/x-1+3 = 2 // - 2

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

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

(1*x^2+2*x^0)/(x^1) = 0 // * x^2

x^1*(1*x^2+2*x^0) = 0

x^1

x^2+2 = 0

x^2+2 = 0

DELTA = 0^2-(1*2*4)

DELTA = -8

DELTA < 0

x in { } 

x belongs to the empty set

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