Từ gt suy ra: \(x+\sqrt{x^2+2019}=\dfrac{2019}{y+\sqrt{y^2+2019}}=\sqrt{y^2+2019}-y\).

Tương tự: \(y+\sqrt{y^2+2019}=\sqrt{x^2+2019}-x\).

Do đó dễ dàng suy ra được: \(x+y=0\).

\(\Rightarrow x=-y\Rightarrow x^{2019}+y^{2019}=x^{2019}+\left(-x\right)^{2019}=0\left(đpcm\right)\).

\(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x+\sqrt{x^2+1}}=y+\sqrt{y^2+1}\\\frac{1}{y+\sqrt{y^2+1}}=x+\sqrt{x^2+1}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-x+\sqrt{x^2+1}=y+\sqrt{y^2+1}\left(1\right)\\-y+\sqrt{y^2+1}=x+\sqrt{x^2+1}\left(2\right)\end{cases}}\)

Cộng vế với vế của (1) và (2) ta có:

\(-2x-2y=0\Leftrightarrow-2\left(x+y\right)=0\Leftrightarrow x+y=0\)

\(\Rightarrow P=x^{2019}+y^{2019}=0\)

Nhân liên hợp cả 2 vế

P=1

Có \(y^2+2019=y^2+xy+yz+zx=y\left(x+y\right)+z\left(x+y\right)=\left(y+z\right)\left(x+y\right)\)

\(x^2+2019=x^2+xy+yz+zx=x\left(x+y\right)+z\left(x+y\right)=\left(x+z\right)\left(x+y\right)\)

\(z^2+2019=z^2+xy+yz+xz=z\left(z+y\right)+x\left(y+z\right)=\left(z+x\right)\left(y+z\right)\)

Có \(P=x\sqrt{\frac{\left(y^2+2019\right)\left(z^2+2019\right)}{x^2+2019}}+y\sqrt{\frac{\left(z^2+2019\right)\left(x^2+2019\right)}{y^2+2019}}+z\sqrt{\frac{\left(x^2+2019\right)\left(y^2+2019\right)}{z^2+2019}}\)

=\(x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(z+y\right)}{\left(x+z\right)\left(y+x\right)}}+y\sqrt{\frac{\left(z+x\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(y+z\right)\left(x+y\right)}}+z\sqrt{\frac{\left(x+z\right)\left(x+y\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)

=\(x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)

=\(x\left|y+z\right|+y\left|x+z\right|+z\left|x+y\right|\)

=\(x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\) (vì x,y,z >0)

= xy+xz+xy+yz+xz+yz

=2(xy+xz+yz)=2.2019(vì xy+xz+yz=2019)

=4038

Vậy P=4038

\(\left(x+\sqrt{x^2+2019}\right)\left(\sqrt{x^2+2019}-x\right)=x^2+2019-x^2=2019\)

\(\Rightarrow\sqrt{x^2+2019}-x=y+\sqrt{y^2+2019}\left(2\right)\)

Tương tự \(\sqrt{y^2+2019}-y=x+\sqrt{x^2+2019}\left(1\right)\)

Lấy (2) - (1) được: -2x = 2y

                       <=> -x = y

                       <=> x + y = 0