(4x-8)/(x^2-4)*(x+2)/(5x^2-25x)/(28)/(x^2-25)

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Solution for (4x-8)/(x^2-4)*(x+2)/(5x^2-25x)/(28)/(x^2-25) equation:


D( x )

x^2-4 = 0

5*x^2-(25*x) = 0

x^2-25 = 0

x^2-4 = 0

x^2-4 = 0

1*x^2 = 4 // : 1

x^2 = 4

x^2 = 4 // ^ 1/2

abs(x) = 2

x = 2 or x = -2

5*x^2-(25*x) = 0

5*x^2-(25*x) = 0

5*x^2-25*x = 0

5*x^2-25*x = 0

DELTA = (-25)^2-(0*4*5)

DELTA = 625

DELTA > 0

x = (625^(1/2)+25)/(2*5) or x = (25-625^(1/2))/(2*5)

x = 5 or x = 0

x^2-25 = 0

x^2-25 = 0

1*x^2 = 25 // : 1

x^2 = 25

x^2 = 25 // ^ 1/2

abs(x) = 5

x = 5 or x = -5

x in (-oo:-5) U (-5:-2) U (-2:0) U (0:2) U (2:5) U (5:+oo)

(((((4*x-8)/(x^2-4))*(x+2))/(5*x^2-(25*x)))/28)/(x^2-25) = 0

(((((4*x-8)/(x^2-4))*(x+2))/(5*x^2-25*x))/28)/(x^2-25) = 0

((4*x-8)*(x+2))/(28*(x^2-4)*(5*x^2-25*x)*(x^2-25)) = 0

5*x^2-25*x = 0

5*x*(x-5) = 0

x-5 = 0 // + 5

x = 5

5*x*(x-5) = 0

((4*x-8)*(x+2))/(5*28*x*(x^2-4)*(x-5)*(x^2-25)) = 0

( 4*x-8 )

4*x-8 = 0 // + 8

4*x = 8 // : 4

x = 8/4

x = 2

( x+2 )

x+2 = 0 // - 2

x = -2

x in { 2}

x in { -2}

x belongs to the empty set

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