1/3-g=(1/3+g)+1/g

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


D( g )

g = 0

g = 0

g = 0

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

1/3-g = g+1/g+1/3 // - g+1/g+1/3

1/3-g-g-(1/g)-(1/3) = 0

1/3-g-g-g^-1-1/3 = 0

-2*g^1-1*g^-1 = 0

(-2*g^2-1*g^0)/(g^1) = 0 // * g^2

g^1*(-2*g^2-1*g^0) = 0

g^1

-2*g^2-1 = 0

-2*g^2-1 = 0

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

DELTA = -8

DELTA < 0

g in { } 

g belongs to the empty set

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