((3x^2-27)/4)*((8x^2)/(9-3x))/((x^2+3x)/6)

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Solution for ((3x^2-27)/4)*((8x^2)/(9-3x))/((x^2+3x)/6) equation:


D( x )

9-(3*x) = 0

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

9-(3*x) = 0

9-(3*x) = 0

9-3*x = 0

9-3*x = 0 // - 9

-3*x = -9 // : -3

x = -9/(-3)

x = 3

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

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

x^2+3*x = 0

x*(x+3) = 0

x+3 = 0 // - 3

x = -3

x*(x+3) = 0

(x*(x+3))/6 = 0

( x+3 )

x+3 = 0 // - 3

x = -3

( x )

x = 0

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

(((3*x^2-27)/4)*((8*x^2)/(9-(3*x))))/((x^2+3*x)/6) = 0

(((3*x^2-27)/4)*((8*x^2)/(9-3*x)))/((x^2+3*x)/6) = 0

(6*8*x^2*(3*x^2-27))/(4*(9-3*x)*(x^2+3*x)) = 0

x^2+3*x = 0

x*(x+3) = 0

x+3 = 0 // - 3

x = -3

x*(x+3) = 0

(6*8*x^2*(3*x^2-27))/(4*x*(9-3*x)*(x+3)) = 0

( 3*x^2-27 )

3*x^2 = 27 // : 3

x^2 = 9

x^2 = 9 // ^ 1/2

abs(x) = 3

x = 3 or x = -3

( 8*x^2 )

8*x^2 = 0 // : 8

x^2 = 0

x = 0

x in { 3}

x in { -3}

x in { 0}

x belongs to the empty set

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