Có \(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)
\(\Leftrightarrow\left(x+\sqrt{x^2+1}\right)\left(x-\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=x-\sqrt{x^2+1}\)
\(\Leftrightarrow\left[x^2-\left(\sqrt{x^2+1}\right)^2\right]\left(y+\sqrt{y^2+1}\right)=x-\sqrt{x^2+1}\)
\(\Leftrightarrow-y-\sqrt{y^2+1}=x-\sqrt{x^2+1}\) (1)
Lại có:\(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)
\(\Leftrightarrow\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)\left(y-\sqrt{y^2+1}\right)=y-\sqrt{y^2+1}\)
\(\Leftrightarrow\left(x+\sqrt{x^2+1}\right)\left[y^2-\left(\sqrt{y^2+1}\right)^2\right]=y-\sqrt{y^2+1}\)
\(\Leftrightarrow-x-\sqrt{x^2+1}=y-\sqrt{y^2+1}\) (2)
Từ (1) và (2) cộng vế với vế có:
\(-\left(y+x\right)-\left(\sqrt{x^2+1}+\sqrt{y^2+1}\right)=x+y-\left(\sqrt{x^2+1}+\sqrt{y^2+1}\right)\)
\(\Leftrightarrow2\left(x+y\right)=0\)
\(\Leftrightarrow x+y=0\) hay S=0
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