Question

Gia sử x, y là hai số thực phân biệt thỏa mãn: \(\frac{1}{x^2+1}+\frac{1}{y^2+1}=\frac{2}{xy+1}\). Hãy tính: S=\(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{2}{xy+1}\)

Answer 1

\(\frac{1}{x^2+1}+\frac{1}{y^2+1}=\frac{2}{xy+1}\) (điều kiện: \(xy\ne-1\))

\(\Leftrightarrow\frac{x^2+y^2+2}{x^2y^2+x^2+y^2+1}=\frac{2}{xy+1}\)

\(\Leftrightarrow\left(xy+1\right)\left(x^2+y^2+2\right)=2x^2y^2+2x^2+2y^2+2\)

\(\Leftrightarrow xy\left(x^2+y^2\right)+2xy+x^2+y^2+2=2x^2y^2+2x^2+2y^2+2\)

\(\Leftrightarrow xy\left(x^2+y^2-2xy\right)+2xy-x^2-y^2=0\)

\(\Leftrightarrow\left(xy-1\right)\left(x-y\right)^2=0\Rightarrow\left[{}\begin{matrix}xy=1\left(l\right)\\x=y\end{matrix}\right.\)

\(\Rightarrow xy=1\)

\(S=\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{2}{xy+1}=\frac{4}{xy+1}=\frac{4}{1+1}=2\)