Question

AL 66 Poli Timisoara 2023

Answer 1

$\it (x+\sqrt{1+x^2})(y+\sqrt{1+y^2})=1\bigg|_{\cdot(x-\sqrt{1+x^2})} \Rightarrow -y-\sqrt{1+y^2}=x-\sqrt{1+x^2} \Rightarrow\\ \\ \\ \Rightarrow x+y=\sqrt{1+x^2}-\sqrt{1+y^2}\ \ \ \ \ (1)\\ \\ \\ (x+\sqrt{1+x^2})(y+\sqrt{1+y^2})=1\bigg|_{\cdot(y-\sqrt{1+y^2})} \Rightarrow -x-\sqrt{1+x^2}=y-\sqrt{1+y^2} \Rightarrow\\ \\ \\ \Rightarrow x+y=\sqrt{1+y^2}-\sqrt{1+x^2}\ \ \ \ \ (2)$

$\it (1),\ (2) \Rightarrow x+y=0 \Rightarrow (x+y)^{2023}=0^{2023}=0\\ \\ R\breve aspunsul\ corect\ este\ b)\ 0$