Đặt \(\left\{{}\begin{matrix}x+\sqrt{1+x^2}=a>0\\y+\sqrt{1+y^2}=b>0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{1+x^2}=a-x\\\sqrt{1+y^2}=b-y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}1+x^2=a^2+x^2-2ax\\1+y^2=b^2+y^2-2by\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{a^2-1}{2a}\\y=\dfrac{b^2-1}{2b}\end{matrix}\right.\) thế vào điều kiện đề bài:
\(\left(\dfrac{a^2-1}{2a}+\sqrt{1+\dfrac{b^4-2b^2+1}{4b^2}}\right)\left(\dfrac{b^2-1}{2b}+\sqrt{1+\dfrac{a^4-2a^2+1}{4a^2}}\right)=1\)
\(\Leftrightarrow\left(\dfrac{a^2-1}{2a}+\dfrac{b^2+1}{2b}\right)\left(\dfrac{b^2-1}{2b}+\dfrac{a^2+1}{2a}\right)=1\)
\(\Leftrightarrow\left(\dfrac{a+b}{2}+\dfrac{a-b}{2ab}\right)\left(\dfrac{a+b}{2}-\dfrac{a-b}{2ab}\right)=1\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{4}-\dfrac{\left(a-b\right)^2}{4a^2b^2}=1=\dfrac{\left(a+b\right)^2-\left(a-b\right)^2}{4ab}\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{4}-\dfrac{\left(a+b\right)^2}{4ab}+\dfrac{\left(a-b\right)^2}{4ab}-\dfrac{\left(a-b\right)^2}{4a^2b^2}=0\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{4}\left(1-\dfrac{1}{ab}\right)+\dfrac{\left(a-b\right)^2}{4ab}\left(1-\dfrac{1}{ab}\right)=0\)
\(\Leftrightarrow\left(1-\dfrac{1}{ab}\right)\left[\dfrac{\left(a+b\right)^2}{4}+\dfrac{\left(a-b\right)^2}{4ab}\right]=0\)
\(\Leftrightarrow1-\dfrac{1}{ab}=0\)
\(\Leftrightarrow ab=1\)
\(\Leftrightarrow\left(x+\sqrt{1+x^2}\right)\left(y+\sqrt{1+y^2}\right)=1\) (đpcm)
