(1/x+2)+(1/x+3)=(2/x^2+5x+6)

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Solution for (1/x+2)+(1/x+3)=(2/x^2+5x+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+1/x+2+3 = 5*x+2/(x^2)+6 // - 5*x+2/(x^2)+6

1/x-(5*x)+1/x-(2/(x^2))-6+2+3 = 0

1/x-5*x+1/x-2*x^-2-6+2+3 = 0

2*x^-1-5*x^1-2*x^-2-1*x^0 = 0

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

x^2*(2*x^1-5*x^3-1*x^2-2*x^0) = 0

x^2

2*x-5*x^3-x^2-2 = 0

{ 1, -1, 2, -2 }

1

x = 1

2*x-5*x^3-x^2-2 = -6

1

-1

x = -1

2*x-5*x^3-x^2-2 = 0

-1

x+1

4*x-5*x^2-2

2*x-5*x^3-x^2-2

x+1

5*x^3+5*x^2

4*x^2+2*x-2

-4*x^2-4*x

-2*x-2

2*x+2

0

4*x-5*x^2-2 = 0

DELTA = 4^2-(-5*(-2)*4)

DELTA = -24

DELTA < 0

x in { -1} 

x = -1

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