(1/x-2)-(3/x+3)=(4/x^2+x-6)

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Solution for (1/x-2)-(3/x+3)=(4/x^2+x-6) 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)

1/x-(3/x)-3-2 = x+4/(x^2)-6 // - x+4/(x^2)-6

1/x-x-(3/x)-(4/(x^2))-3-2+6 = 0

1/x-x-3*x^-1-4*x^-2-3-2+6 = 0

1*x^0-1*x^1-2*x^-1-4*x^-2 = 0

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

x^2*(1*x^2-1*x^3-2*x^1-4*x^0) = 0

x^2

x^2-x^3-2*x-4 = 0

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

1

x = 1

x^2-x^3-2*x-4 = -6

1

-1

x = -1

x^2-x^3-2*x-4 = 0

-1

x+1

2*x-x^2-4

x^2-x^3-2*x-4

x+1

x^3+x^2

2*x^2-2*x-4

-2*x^2-2*x

-4*x-4

4*x+4

0

2*x-x^2-4 = 0

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

DELTA = -12

DELTA < 0

x in { -1} 

x = -1

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