(4x^2-4x+8)/(1-x^2)

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Solution for (4x^2-4x+8)/(1-x^2) equation: 


D( x )

1-x^2 = 0

1-x^2 = 0

1-x^2 = 0

-1*x^2 = -1 // : -1

x^2 = 1

x^2 = 1 // ^ 1/2

abs(x) = 1

x = 1 or x = -1

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

(4*x^2-(4*x)+8)/(1-x^2) = 0

(4*x^2-4*x+8)/(1-x^2) = 0

4*x^2-4*x+8 = 0

4*(x^2-x+2) = 0

x^2-x+2 = 0

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

DELTA = -7

DELTA < 0

4 = 0

4/(1-x^2) = 0

x belongs to the empty set

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