Đáp án: $A=2019$
Giải thích các bước giải:
Ta có:
$(x+\sqrt{x^2+2019})(y+\sqrt{y^2+2019})=2019$
$\to (x+\sqrt{x^2+2019})(y+\sqrt{y^2+2019})=(x^2+2019)-x^2$
$\to (x+\sqrt{x^2+2019})(y+\sqrt{y^2+2019})=(\sqrt{x^2+2019}-x)(\sqrt{x^2+2019}+x)$
$\to y+\sqrt{y^2+2019}=\sqrt{x^2+2019}-x$ vì $x+\sqrt{x^2+2019}>0\quad\forall x>0$
$\to x+y=\sqrt{x^2+2019}-\sqrt{y^2+2019}(1)$
Tương tự ta có:
$(x+\sqrt{x^2+2019})(y+\sqrt{y^2+2019})=2019$
$\to (x+\sqrt{x^2+2019})(y+\sqrt{y^2+2019})=(y^2+2019)-y^2$
$\to (x+\sqrt{x^2+2019})(y+\sqrt{y^2+2019})=(\sqrt{y^2+2019}-y)(\sqrt{y^2+2019}+y)$
$\to x+\sqrt{x^2+2019}=\sqrt{y^2+2019}-y$ vì $y+\sqrt{y^2+2019}>0\quad\forall y>0$
$\to x+y=\sqrt{y^2+2019}-\sqrt{x^2+2019}(2)$
Cộng vế với vế của $(1),(2)$
$\to 2(x+y)=0$
$\to x+y=0$
$\to x=-y$
$\to x^{2019}=-y^{2019}$
$\to A=2019$
