Vì \(\sqrt{\left(x+y\right)^2}=\left|x+y\right|\ge0\forall x;y\)

\(\sqrt{\left(y-2005\right)^2}=\left|y-2005\right|\ge0\forall y\)

\(\Rightarrow\sqrt{\left(x+y\right)^2}+\sqrt{\left(y-2005\right)^2}\ge0\forall x;y\)

Mà \(\sqrt{\left(x+y\right)^2}+\sqrt{\left(y-2005\right)^2}< 0\Rightarrow x;y\in\varphi\)

Vậy \(x;y\in\varphi\)

1)    \(\left(x+\sqrt{x^2+\sqrt{2005}}\right)\left(\sqrt{x^2+\sqrt{2005}}-x\right)=\sqrt{2005}\)

Kết hợp với giả thiết ta được:

     \(\sqrt{x^2+\sqrt{2005}}-x=y+\sqrt{y^2+\sqrt{2005}}\)

suy ra: đpcm

2)     \(\left(x+\sqrt{x^2+\sqrt{2005}}\right)\left(y+\sqrt{y^2+\sqrt{2005}}\right)=\sqrt{2005}\)

Ta có:  \(\hept{\begin{cases}\left(x+\sqrt{x^2+\sqrt{2005}}\right)\left(\sqrt{x^2+\sqrt{2005}}-x\right)=\sqrt{2005}\\\left(y+\sqrt{y^2+\sqrt{2005}}\right)\left(\sqrt{y^2+\sqrt{2005}}-y\right)=\sqrt{2005}\end{cases}}\)

Kết hợp với giả thiết ta có:

\(\hept{\begin{cases}\sqrt{x^2+\sqrt{2005}}-x=y+\sqrt{y^2+\sqrt{2005}}\\\sqrt{y^2+\sqrt{2005}}-y=x+\sqrt{x^2+\sqrt{2005}}\end{cases}}\)

suy ra:  \(x+y=-\left(x+y\right)\)

\(\Rightarrow\)\(S=x+y=0\)

\(a^2=b+4010\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2+4010\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz=x^2+y^2+z^2+4010\)

\(\Rightarrow2xy+2yz+2xz=4010\Rightarrow xy+yz+xz=2005\)

\(x\sqrt{\frac{\left(2015+y^2\right)\left(2005+z^2\right)}{\left(2005+x^2\right)}}=x\sqrt{\frac{\left(xz+yz+xy+y^2\right)\left(xy+xz+yz+z^2\right)}{\left(xy+yz+x^2+xz\right)}}\)

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

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

tương tự : \(y\sqrt{\frac{\left(2015+x^2\right)\left(2015+z^2\right)}{2015+y^2}}=xy+yz;z\sqrt{\frac{\left(2005+x^2\right)\left(2005+y^2\right)}{2015+z^2}}=xz+yz\)

\(\Rightarrow M=xy+xz+xy+yz+xz+yz=2\left(xy+yz+xz\right)=2\cdot2005=4010\)