soluzione Calcolare x2 x2 + 2xy + y2 · x2 + y2 x2y2 + 2x2 x3 + 3x2y + 3xy2 + y3 · ( 1 x + 1 y ) =   =   x2 x2 + 2xy + y2 · x2 + y2 x2y2 + 2x2 x3 + 3x2y + 3xy2 + y3 · y + x xy   =     =   x2 (x + y)2 · x2 + y2 x2y2 + 2x2 (x + y)3 · x + y xy   =     =   x2 + y2 y2(x + y)2 + 2x y(x + y)2   =     =   x2 + y2 + 2xy y2(x + y)2   =     =   (x + y)2 y2(x + y)2   =     =   1 y2