Đáp án:
P=2020
Giải thích các bước giải:
$\quad\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}=\dfrac{2}{xy+1}\\ \rightarrow (\dfrac{1}{x^2+1}-\dfrac{1}{xy+1})+(\dfrac{1}{y^2+1}-\dfrac{1}{xy+1})=0\\ \rightarrow \dfrac{xy+1-x^2-1}{(x^2+1)(xy+1)}+ \dfrac{xy+1-y^2-1}{(y^2+1)(xy+1)}=0\\ \rightarrow \dfrac{-x(x-y)}{(x^2+1)(xy+1)}+\dfrac{y(x-y)}{(y^2+1)(xy+1)}=0\\ \rightarrow\dfrac{x-y}{xy+1}(\dfrac{-x}{x^2+1}+\dfrac{y}{y^2+1})=0\\ \rightarrow \dfrac{x-y}{xy+1}\dfrac{-x(y^2+1)+y(x^2+1)}{(x^2+1)(y^2+1)}=0 \\ \rightarrow\dfrac{x-y}{xy+1}\dfrac{(x-y)(xy-1)}{(x^2+1)(y^2+1)}=0\\ \rightarrow xy=1 \quad\text{ Do x,y phân biệt}\\ \rightarrow P=1010(\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}+\dfrac{2}{xy+1})=1010.\dfrac{4}{xy+1}=2020$
