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

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


D( y )

y-1 = 0

(y+1)/(2*y-2) = 0

2*y-2 = 0

y-1 = 0

y-1 = 0

y-1 = 0 // + 1

y = 1

(y+1)/(2*y-2) = 0

(y+1)/(2*y-2) = 0

y+1 = 0 // - 1

y = -1

2*y-2 = 0

2*y-2 = 0

2*y-2 = 0 // + 2

2*y = 2 // : 2

y = 2/2

y = 1

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

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

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

( 2*y-2 )

2*y-2 = 0 // + 2

2*y = 2 // : 2

y = 2/2

y = 1

( y^2-1 )

1*y^2 = 1 // : 1

y^2 = 1

y^2 = 1 // ^ 1/2

abs(y) = 1

y = 1 or y = -1

y in { 1} 

y in { -1} 

y belongs to the empty set

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