(-2/x-3)+(3/x+3)=(-12/x^2-9)

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Solution for (-2/x-3)+(3/x+3)=(-12/x^2-9) 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-3+3 = -12/(x^2)-9 // + -12/(x^2)-9

3/x-2/x-(-12/(x^2))-3+3+9 = 0

3/x-2/x+12*x^-2-3+3+9 = 0

x^-1+12*x^-2+9 = 0

t_1 = x^-1

12*t_1^2+1*t_1^1+9 = 0

12*t_1^2+t_1+9 = 0

DELTA = 1^2-(4*9*12)

DELTA = -431

DELTA < 0

x belongs to the empty set

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