(x^3+3x^2)/(2x)*(5x^3)/(x^2+5x+6)

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Solution for (x^3+3x^2)/(2x)*(5x^3)/(x^2+5x+6) equation:


D( x )

x^2+5*x+6 = 0

2*x = 0

x^2+5*x+6 = 0

x^2+5*x+6 = 0

x^2+5*x+6 = 0

DELTA = 5^2-(1*4*6)

DELTA = 1

DELTA > 0

x = (1^(1/2)-5)/(1*2) or x = (-1^(1/2)-5)/(1*2)

x = -2 or x = -3

2*x = 0

2*x = 0

2*x = 0 // : 2

x = 0

x in (-oo:-3) U (-3:-2) U (-2:0) U (0:+oo)

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

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

x^3+3*x^2 = 0

x^2*(x+3) = 0

x+3 = 0 // - 3

x = -3

x^2*(x+3) = 0

x^2+5*x+6 = 0

x^2+5*x+6 = 0

DELTA = 5^2-(1*4*6)

DELTA = 1

DELTA > 0

x = (1^(1/2)-5)/(1*2) or x = (-1^(1/2)-5)/(1*2)

x = -2 or x = -3

(x+3)*(x+2) = 0

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

( 5*x^3 )

5*x^3 = 0 // : 5

x^3 = 0

x = 0

( x+3 )

x+3 = 0 // - 3

x = -3

( x^2 )

1*x^2 = 0 // : 1

x^2 = 0

x = 0

x in { 0}

x in { -3}

x in { 0}

x belongs to the empty set

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