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

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Solution for (3x-3)/(x+2)/(x^2-1)/(x^2+5x+6) equation:


D( x )

x^2+5*x+6 = 0

x^2-1 = 0

x+2 = 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

x^2-1 = 0

x^2-1 = 0

1*x^2 = 1 // : 1

x^2 = 1

x^2 = 1 // ^ 1/2

abs(x) = 1

x = 1 or x = -1

x+2 = 0

x+2 = 0

x+2 = 0 // - 2

x = -2

x in (-oo:-3) U (-3:-2) U (-2:-1) U (-1:1) U (1:+oo)

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

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

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

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

3*x-3 = 0 // + 3

3*x = 3 // : 3

x = 3/3

x = 1

x in { 1}

x belongs to the empty set

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