1/(2x)+3/(3x^2+12x)=3/x

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Solution for 1/(2x)+3/(3x^2+12x)=3/x equation:


D( x )

x = 0

2*x = 0

3*x^2+12*x = 0

x = 0

x = 0

2*x = 0

2*x = 0

2*x = 0 // : 2

x = 0

3*x^2+12*x = 0

3*x^2+12*x = 0

3*x^2+12*x = 0

DELTA = 12^2-(0*3*4)

DELTA = 144

DELTA > 0

x = (144^(1/2)-12)/(2*3) or x = (-144^(1/2)-12)/(2*3)

x = 0 or x = -4

x in (-oo:-4) U (-4:0) U (0:+oo)

3/(3*x^2+12*x)+1/(2*x) = 3/x // - 3/x

3/(3*x^2+12*x)+1/(2*x)-(3/x) = 0

3/(3*x^2+12*x)+1/(2*x)-3*x^-1 = 0

3/(3*x^2+12*x)+1/(2*x)-3/x = 0

3*x^2+12*x = 0

3*x^2+12*x = 0

3*x*(x+4) = 0

x+4 = 0 // - 4

x = -4

3*x*(x+4) = 0

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

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

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

3*x^3-18*x^3+18*x^2-72*x^2 = 0

-15*x^3-54*x^2 = 0

-15*x^3-54*x^2 = 0

-3*x^2*(5*x+18) = 0

5*x+18 = 0 // - 18

5*x = -18 // : 5

x = -18/5

-3*x^2*(x+18/5) = 0

(-3*x^2*(x+18/5))/(2*3*x*x*x*(x+4)) = 0

(-3*x^2*(x+18/5))/(2*3*x*x*x*(x+4)) = 0 // * 2*3*x*x*x*(x+4)

-3*x^2*(x+18/5) = 0

( -3*x^2 )

-3*x^2 = 0 // : -3

x^2 = 0

x = 0

( x+18/5 )

x+18/5 = 0 // - 18/5

x = -18/5

x in { 0}

x = -18/5

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