soluzione Calcolare [(1 + x2+1 x : x2-1 x ) : (2 - x2+2 x : x2-1 x ) -( x x+2 + 1 x-2 )] · x2-x-2 x2- 1   =     =   [(1 + x2+1 x · x x2-1 ) : (2 - x2+2 x · x x2-1 ) -( x x+2 + 1 x-2 )] · x2-x-2 x2- 1   =     =   [(1 + x2+1 x2-1 ) : (2 - x2+2 x2-1 ) -( x x+2 + 1 x-2 )] · x2-x-2 x2- 1   =     =   [(1 + x2+1 (x - 1)(x + 1) ) : (2 - x2+2 (x - 1)(x + 1) ) -( x x+2 + 1 x-2 )] · x2-x-2 x2- 1   =     =   [ x2 - 1 + x2 + 1 (x - 1)(x + 1) : 2(x2 - 1) - (x2 + 2) (x - 1)(x + 1) - x - 2 + x + 2 (x - 2)(x + 2) ] · x2-x-2 x2- 1   =     =   [ x2 - 1 + x2 + 1 (x - 1)(x + 1) : 2x2 - 2 - x2 - 2 (x - 1)(x + 1) - x - 2 + x + 2 (x - 2)(x + 2) ] · x2-x-2 x2- 1   =     =   [ 2x2 (x - 1)(x + 1) : x2 - 4 (x - 1)(x + 1) - 2x (x - 2)(x + 2) ] · x2-x-2 x2- 1   =     =   [ 2x2 (x - 1)(x + 1) · (x - 1)(x + 1) x2 - 4 - 2x (x - 2)(x + 2) ] · x2-x-2 x2- 1   =     =   [ 2x2 x2 - 4 - 2x (x - 2)(x + 2) ] · x2-x-2 x2- 1   =     =   [ 2x2 (x - 2)(x + 2) - 2x (x - 2)(x + 2) ] · x2-x-2 x2- 1   =     =   2x2 - 2x (x - 2)(x + 2) · x2-x-2 x2- 1   =     =   2x(x - 1) (x - 2)(x + 2) · (x - 2)(x + 1) (x - 1)(x + 1)   =     =   2x x + 2