((x+4)/(x^2-5x+6))/((x^2-16)/(x+3))

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Solution for ((x+4)/(x^2-5x+6))/((x^2-16)/(x+3)) equation:


D( x )

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

(x^2-16)/(x+3) = 0

x+3 = 0

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

x = 3 or x = 2

(x^2-16)/(x+3) = 0

(x^2-16)/(x+3) = 0

1*x^2 = 16 // : 1

x^2 = 16

x^2 = 16 // ^ 1/2

abs(x) = 4

x = 4 or x = -4

x+3 = 0

x+3 = 0

x+3 = 0 // - 3

x = -3

x in (-oo:-4) U (-4:-3) U (-3:2) U (2:3) U (3:4) U (4:+oo)

((x+4)/(x^2-(5*x)+6))/((x^2-16)/(x+3)) = 0

((x+4)/(x^2-5*x+6))/((x^2-16)/(x+3)) = 0

((x+4)*(x+3))/((x^2-5*x+6)*(x^2-16)) = 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 = (5-1^(1/2))/(1*2)

x = 3 or x = 2

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

((x+4)*(x+3))/((x-2)*(x-3)*(x^2-16)) = 0

( x+3 )

x+3 = 0 // - 3

x = -3

( x+4 )

x+4 = 0 // - 4

x = -4

x in { -3}

x in { -4}

x belongs to the empty set

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