Ta có: \(\left(\sqrt{a^2+1}-a\right)\left(\sqrt{b^2+1}-b\right)=1\)
Dễ thấy: \(\left\{{}\begin{matrix}\left(\sqrt{a^2+1}-a\right)\ne0\\\left(\sqrt{b^2+1}-b\right)\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(\sqrt{a^2+1}-a\right)\left(\sqrt{b^2+1}-b\right)\left(\sqrt{a^2+1}+a\right)=\sqrt{a^2+1}+a\\\left(\sqrt{a^2+1}-a\right)\left(\sqrt{b^2+1}-b\right)\left(\sqrt{b^2+1}+b\right)=\sqrt{b^2+1}+b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{b^2+1}-b=\sqrt{a^2+1}+a\left(1\right)\\\sqrt{a^2+1}-a=\sqrt{b^2+1}+b\left(2\right)\end{matrix}\right.\)
Lấy (1) + (2) vế theo vế ta được
\(2\left(a+b\right)=0\)
\(\Rightarrow a+b=0\)
