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