$$A(x)=1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{x}}}$$
$$B(x)=1+\dfrac{x}{1+\dfrac{x}{1+\dfrac{1}{x}}}$$
SOLUCIÓN.
$A(x)=1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{x}}}=1+\dfrac{1}{1+\dfrac{1}{\dfrac{x+1}{x}}}=1+\dfrac{1}{1+\dfrac{x}{x+1}}=$
$=1+\dfrac{1}{\dfrac{2x+1}{x+1}}=1+\dfrac{x+1}{2x+1}=\dfrac{2x+1+x+1}{2x+1}=\dfrac{3x+2}{2x+1}$
$B(x)=1+\dfrac{x}{1+\dfrac{x}{1+\dfrac{1}{x}}}=1+\dfrac{x}{1+\dfrac{x}{\dfrac{x+1}{x}}}=1+\dfrac{x}{1+\dfrac{x^2}{x+1}}=$
$=1+\dfrac{x}{\dfrac{x^2+x+1}{x+1}}=1+\dfrac{x\,(x+1)}{x^2+x+1}=\dfrac{(x^2+x+1)+(x^2+x)}{x^2+x+1}=$
$=\dfrac{2\,x^2+2\,x+1}{x^2+x+1}$
Entonces, $$A(x) \cdot B(x) = \dfrac{3x+2}{2x+1} \cdot \dfrac{2\,x^2+2\,x+1}{x^2+x+1}=$$ $$=\dfrac{(3x+2)(2\,x^2+2\,x+1)}{(2x+1)(x^2+x+1)}=\dfrac{6\,x^3+10\,x^2+7\,x+2}{2\,x^3+3\,x^2+3\,x+1}$$
$\square$
[autoría]
