Question

Giả sử (\(\sqrt{a^2+1}\)- a)(\(\sqrt{b^2+1}-b\))=1. Tính a+b

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Answer 1

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

\(2\left(a+b\right)=0\)

\(\Rightarrow a+b=0\)