Question

Acolo la fractie e 4x+4y/4+xy

Answer 1

$x*y = \frac{\Large 4x+4y}{\Large 4+xy}\\\\f(x)=\frac{\Large 2(x-1)}{\Large x+1}\\\\ f(xy) = \frac{\Large 2(xy - 1)}{\Large xy + 1} = \boxed{\frac{\Large 2xy - 2}{\Large xy+1}}\\\\ f(x)*f(y) = \frac{\Large 4f(x) + 4f(y)}{\Large 4+f(x)f(y)} = \frac{\Large 4\Big(f(x) + f(y)\Big)}{\Large 4 + f(x)f(y)} = \frac{\Large 4\Big(\frac{\Large 2(x-1)}{\Large x+1} + \frac{\Large 2(y-1)}{\Large y+1} \Big)}{\Large 4+\frac{\Large 2(x-1)}{\Large x+1} \frac{\Large 2(y-1)}{\Large y+1}} = \frac{\Large 4\Big(\frac{\Large 2(x-1)}{\Large x+1} + \frac{\Large 2(y-1)}{\Large y+1} \Big)}{\Large 4\Big(1 + \frac{\Large (x-1)(y-1)}{\Large (x+1)(y+1)} \Big)} = \frac{\Large \frac{\Large 2(x-1)}{\Large x+1} + \frac{\Large 2(y-1)}{\Large y+1} }{1 + \frac{\Large (x-1)(y-1)}{\Large (x+1)(y+1)} } = \frac{\Large\frac{\Large 2(x-1)(y+1) + 2(y-1)(x+1)}{\Large (x+1)(y+1)}}{\frac{\Large (x+1)(y+1) + (x-1)(y-1) }{\Large (x+1)(y+1)}} = \frac{\Large 2(xy +x -y -1) + 2(xy+y -x -1)}{\Large xy + x + y + 1 + xy -x -y +1} = \frac{\Large 2xy + 2x -2y - 2 + 2xy + 2y -2x -2}{\Large 2xy + 2} = \frac{\Large 4xy - 4}{\Large 2xy + 2} = \frac{\Large 4(xy-1)}{\Large 2(xy+1)} = \frac{\Large 2(xy-1)}{\Large xy+1} = \boxed{\frac{\Large 2xy - 2}{\Large xy+1}, \forall x,y \in (0, +\infty)}$