Đáp án: $S=2$
Giải thích các bước giải:
$\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}=\dfrac{2}{xy+1}$
$\rightarrow (\dfrac{1}{x^2+1}-\dfrac{1}{xy+1})+(\dfrac{1}{y^2+1}-\dfrac{1}{xy+1})=0$
$\rightarrow \dfrac{xy+1-x^2-1}{(x^2+1)(xy+1)}+\dfrac{xy+1-y^2-1}{(y^1+1)(xy+1)}=0$
$\rightarrow \dfrac{x(y-x)}{(x^2+1)(xy+1)}+\dfrac{y(x-y)}{(y^1+1)(xy+1)}=0$
$\rightarrow (x-y)(\dfrac{-x}{(x^2+1)(xy+1)}+\dfrac{y}{(y^1+1)(xy+1)})=0$
$\rightarrow (x-y)(\dfrac{-x(y^2+1)}{(y^2+1)(x^2+1)(xy+1)}+\dfrac{y(x^2+1)}{(x^2+1)(y^1+1)(xy+1)})=0$
$\rightarrow (x-y)(\dfrac{-x(y^2+1)+y(x^2+1)}{(x^2+1)(y^1+1)(xy+1)})=0$
$\rightarrow (x-y)\dfrac{xy(x-y)-(x-y)}{(x^2+1)(y^1+1)(xy+1)}=0$
$\rightarrow (x-y)\dfrac{(xy-1)(x-y)}{(x^2+1)(y^1+1)(xy+1)}=0$
$\rightarrow \dfrac{(xy-1)(x-y)^2}{(x^2+1)(y^1+1)(xy+1)}=0$
$\rightarrow xy=1$ do $x\ne y$
$\rightarrow \dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}=\dfrac{2}{1+1}=1$
$\rightarrow S=1+\dfrac{2}{1+1}=2$
