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

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


D( x )

x^2-1 = 0

(x+1)/(x-1) = 0

x-1 = 0

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+1)/(x-1) = 0

(x+1)/(x-1) = 0

x+1 = 0 // - 1

x = -1

x-1 = 0

x-1 = 0

x-1 = 0 // + 1

x = 1

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

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

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

( x-1 )

x-1 = 0 // + 1

x = 1

( x^2+1 )

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

x^2 = -1

x in { 1} 

x belongs to the empty set

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